[编程题]二次方程计算器
提供两种方法：
第一种方法：只适用本题的情况，第二种则通用
package com.speical.first;
import java.text.DecimalFormat;
import java.util.Scanner;
/**
* 二元一次方程组解法
*
*
* @author special
* @date 2018年1月2日 下午3:24:01
*/
public class Pro90 {
static DecimalFormat df = new DecimalFormat("0.00");
/**
* 解析从fromIndex 到 toIndex(不包括）的数
* 如果num = 0 且 toIndex不是末尾，那么说明该数
* 是未知数的系数且为1（因为1会省略不写）
* [@param fromIndex
* @param toIndex
* @param expression
* @return](/profile/547241) */
public static int parseInt(int fromIndex, int toIndex, String expression){
int num = 0;
boolean flag = true; //记录正负
if(expression.charAt(fromIndex) == '-'){
flag = false;
fromIndex++;
}else if(expression.charAt(fromIndex) == '+'){
flag = true;
fromIndex++;
}
while(fromIndex < toIndex){
num = num * 10 + (expression.charAt(fromIndex) - '0');
fromIndex++;
}
if(num == 0 && toIndex < expression.length()) num = 1; //如果是0且是未知数的系数，那么为1
return flag ? num : -num;
}
/**
* 知道了3个系数，那么很容易求答案了
* 注意该题的答案保留2位小数不用四舍五入
* 该死的，format会自动四舍五入
* 你若是不想四舍五入，必须借助DecimalFormat类
* @param a
* @param b
* @param c
*/
public static void caculate(int a, int b, int c){
double temp = b * b - 4 * a * c;
if(temp < 0){
System.out.println("No Solution");
}else{
temp = Math.sqrt(temp);
double result1 = ((-b) + temp) / (2 * a);
double result2 = ((-b) - temp) / (2 * a);
if(result1 > result2){
double max = result1;
result1 = result2;
result2 = max;
}
System.out.println(df.format(result1) + " " + df.format(result2));
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner input = new Scanner(System.in);
while(input.hasNext()){
/**
* 这个表达式有个特点
* 就是一次项等号左边和右边都会出现
* 一次项若左边没有，代表这个方程没有一次项了
* 常数项只出现等号右边
*/
String expression = input.nextLine();
int fromIndex = 0, toIndex; //2次项的系数开头索引为0
toIndex = expression.indexOf('x'); //结尾索引为第一次出现x的位置
int a = parseInt(fromIndex, toIndex,expression); //解析出a
fromIndex = expression.indexOf('^') + 2; //一次项的系数开始位置
toIndex = expression.indexOf('x',fromIndex); //一次项的系数结束的位置
int b = 0, c = 0;
if(toIndex != -1){ //注意可能这个二元方程等号左边没有一次项，所以要判断一下
b = parseInt(fromIndex, toIndex, expression); //如果有一次项，解析出来，这里指的是等号左边的
fromIndex = expression.indexOf('=') + 1; //等号右边一次项开始的位置
toIndex = expression.indexOf('x', fromIndex); //等号右边一次项结束的位置
if(toIndex != -1){ //注意可能这个二元方程等号右边没有一次项，所以要判断一下
b -= parseInt(fromIndex, toIndex, expression); //如果等号右边有一次项，解析出来，并减去它
fromIndex = toIndex + 1; //指向常数项开始的位置
}
}else{
fromIndex = expression.indexOf('=') + 1; //若果没有一次项，那么直接找出等号右边的常数项开始的位置
}
toIndex = expression.length(); //常数项结尾的位置，也就是表达式末尾
c = -parseInt(fromIndex, toIndex, expression); // -c
caculate(a,b,c); //计算即可
}
}
}
第二种
package com.speical.improve;
import java.text.DecimalFormat;
import java.util.Scanner;
/**
* 二元一次方程
*
* 分别对等号两边的字符串进行解析，求出各系数
* @author special
* @date 2018年1月2日 下午5:30:22
*/
public class Pro90Improve1 {
static DecimalFormat df = new DecimalFormat("0.00");
static int[] nums = new int[3];
private static boolean isNum(char ch){
return ch >= '0' && ch <= '9';
}
public static void getNum(int fromIndex, int toIndex, String expression, boolean flag){
int num = 0;
char ch;
boolean positive = true;
int[] temp = new int[3];
for(int i = fromIndex; i < toIndex; i++){
ch = expression.charAt(i);
if(isNum(ch)){
while(i < toIndex && isNum(expression.charAt(i))){
num = num * 10 + expression.charAt(i) - '0';
i++;
}
if(i < toIndex && expression.charAt(i) != 'x' || i == toIndex){
temp[2] = positive ? num : -num;
num = 0;
positive = true;
}
i--; //记得减一，因为循环体的最后也有i++
}else if(ch == 'x'){
if(num == 0){
num = 1;
}
if(i < toIndex - 1 && expression.charAt(i + 1) == '^'){
temp[0] = positive ? num : -num;
i += 2;
}else {
temp[1] = positive ? num : -num;
}
num = 0;
positive = true;
}
else if(ch == '-'){
positive = false;
}
}
for(int i = 0; i < 3; i++){
nums[i] = flag ? temp[i] : nums[i] - temp[i];
}
}
/**
* 知道了3个系数，那么很容易求答案了
* 注意该题的答案保留2位小数不用四舍五入
* 该死的，format会自动四舍五入
* 你若是不想四舍五入，必须借助DecimalFormat类
* @param a
* @param b
* @param c
*/
public static void caculate(int a, int b, int c){
double temp = b * b - 4 * a * c;
if(temp < 0){
System.out.println("No Solution");
}else{
temp = Math.sqrt(temp);
double result1 = ((-b) + temp) / (2 * a);
double result2 = ((-b) - temp) / (2 * a);
if(result1 > result2){
double max = result1;
result1 = result2;
result2 = max;
}
System.out.println(df.format(result1) + " " + df.format(result2));
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner input = new Scanner(System.in);
while(input.hasNext()){
String expression = input.nextLine();
int index = expression.indexOf('=');
getNum(0, index, expression, true); //等号左边处理
getNum(index + 1, expression.length(), expression, false); //等号右边处理
caculate(nums[0],nums[1],nums[2]); //计算即可
}
}
}
编辑于 2018-01-02 19:45:24
回复(0)
思路是以'+','=','-'为界，把方程分成每一项，比如"x^2+x=3x+4"分成"x^2","x","3x","4"，若子串有x且有^是二次项，二次项系数加(前面是'+')或减(前面是'-')对应的值（若省略则是1，比如x^2省略前面的常数系数1），子串有x无^是一次项，无x且不空是常数项。另外越过'='相当于变号，把系数全取相反数。
没通过是因为这个用例，0确实是一个根，你这-0什么意思？代码（c++11）：
没通过是因为这个用例，0确实是一个根，你这-0什么意思？代码（c++11）：#include<string>
#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
int main() {
int a, b, c, d, p, q, i, len;//系数abc，判别d
string eqt, sub; // eqt is equation, sub is substring.
while (cin>> eqt) {
a = b = c = p = 0;
len = eqt.length();
for (i=0; i<len+1; ++i)
if (i==len || eqt[i]=='+' || eqt[i]=='=' || eqt[i]=='-') {
sub = eqt.substr(p, i-p);//获得子串并分析
if ((q=sub.find('x')) != string::npos) {//有x是一次或二次项
if (sub.find('^') != string::npos)//有^是二次项，处理a
a += (eqt[p-1]=='-'?-1:1)*(q?stoi(sub.substr(0, q)):1);
else b += (eqt[p-1]=='-'?-1:1)*(q?stoi(sub.substr(0, q)):1);
} else if (sub != "")//没x且不是空则是常数项
c += (eqt[p-1]=='-'?-1:1)*stoi(sub);
if (eqt[i] == '=') { //跨过等号相当于移项，要变号
a = -a; b = -b; c = -c;
}
p = i+1;
}
if (a < 0) { //为保证更小的根先输出，全部变号使a为正且不影响正解
a = -a; b = -b; c = -c;
}
d = b*b-4*a*c; //根的判别式
if (d < 0)
printf("No Solution\n");
else printf("%.2f %.2f\n", (-sqrt(double(d))-double(b))/(2*double(a)),
(sqrt(double(d))-double(b))/(2*double(a)));
}
return 0;
}
发表于 2018-02-17 13:22:20
回复(14)
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <ctype.h>
#include <vector>
#include <cmath>
using namespace std;
int a = 0, b = 0, c = 0;
void get_coe(string str) {
int len = str.length();
// 为了最后时[i+1]操作不会报错
str.append(1, '+');
int t = 0;
int isNegative = 1; // 是否为负数
int afterEqual = 1; // 是否在等号右边
for(int i = 0; i < len; i++) {
if(str[i] == '=') {
afterEqual = -1;
}
if(isdigit(str[i])) {
t = t*10 + (str[i]-'0');
} else if(str[i] == '-') {
isNegative = -1;
} else if(str[i] == '^') {
i++;
} else {
if(str[i] == 'x' && str[i+1] == '^') {
if(t == 0) t = 1;
a += t*isNegative*afterEqual;
} else if(str[i] == 'x') {
if(t == 0) t = 1;
b += t*isNegative*afterEqual;
} else {
c += t*isNegative*afterEqual;
}
t = 0;
isNegative = 1;
}
if(i == len-1 && isdigit(str[i])) {
c += t*isNegative*afterEqual;
break;
}
}
}
int main()
{
string str;
while(cin>>str) {
get_coe(str);
int d = b*b-4*a*c;
if(d < 0) {
cout<<"No Solution"<<endl;
} else {
double t = sqrt(d);
double x1 = (-1*b - t)/(2*a);
double x2 = (-1*b + t)/(2*a);
printf("%.2lf %.2lf\n", min(x1, x2), max(x1, x2));
}
a = 0; b = 0; c = 0;
}
return 0;
}
#include <string>
#include <cstring>
#include <algorithm>
#include <ctype.h>
#include <vector>
#include <cmath>
using namespace std;
int a = 0, b = 0, c = 0;
void get_coe(string str) {
int len = str.length();
// 为了最后时[i+1]操作不会报错
str.append(1, '+');
int t = 0;
int isNegative = 1; // 是否为负数
int afterEqual = 1; // 是否在等号右边
for(int i = 0; i < len; i++) {
if(str[i] == '=') {
afterEqual = -1;
}
if(isdigit(str[i])) {
t = t*10 + (str[i]-'0');
} else if(str[i] == '-') {
isNegative = -1;
} else if(str[i] == '^') {
i++;
} else {
if(str[i] == 'x' && str[i+1] == '^') {
if(t == 0) t = 1;
a += t*isNegative*afterEqual;
} else if(str[i] == 'x') {
if(t == 0) t = 1;
b += t*isNegative*afterEqual;
} else {
c += t*isNegative*afterEqual;
}
t = 0;
isNegative = 1;
}
if(i == len-1 && isdigit(str[i])) {
c += t*isNegative*afterEqual;
break;
}
}
}
int main()
{
string str;
while(cin>>str) {
get_coe(str);
int d = b*b-4*a*c;
if(d < 0) {
cout<<"No Solution"<<endl;
} else {
double t = sqrt(d);
double x1 = (-1*b - t)/(2*a);
double x2 = (-1*b + t)/(2*a);
printf("%.2lf %.2lf\n", min(x1, x2), max(x1, x2));
}
a = 0; b = 0; c = 0;
}
return 0;
}
发表于 2018-03-20 19:28:56
回复(0)
思路：逐个解析字符串分析出系数，用到一点小技巧，利用堆栈存储数据和操作符，利用getch和ungetch来实现流式操作。
/* quadratic equation computation
* eg: input: x^2+x=3x+4
* output: -1.24 3.24
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <ctype.h>
#include <stack>
#include <algorithm>
using namespace std;
#define NUMBER '0'
#define MAXOP (100)
char str[MAXOP];
int pos = 0;
int getch()
{
if (pos < strlen(str))
return str[pos++];
return EOF;
}
void ungetch()
{
pos--;
}
int getop(char s[])
{
int c, i = 0;
s[0] = c = getch();
s[1] = '\0';
if (!isdigit(c))
return c;
while (isdigit(s[++i] = c = getch()))
;
s[i] = '\0';
if (c != EOF)
ungetch();
return NUMBER;
}
int main()
{
while (scanf("%s", str) != EOF) {
char s[10]; /* max coefficient bits is 10 */
int type, a=0, b=0, c=0, left=1;
stack<char> opr;
stack<int> opd;
pos = 0;
while ((type = getop(s)) != EOF) {
switch(type) {
case NUMBER:
int t; char ch;
t = left * atoi(s);
if (!opr.empty() && (ch = opr.top()) == '-') {
t = -t;
opr.pop();
}
opd.push(t);
break;
case '+': case '-':
opr.push(type);
break;
case '=':
left = -1;
break;
case '^':
if (opd.empty()) {
a += 1;
} else {
a += opd.top();
opd.pop();
}
getch(); /*eat following '2'*/
break;
case 'x':
if (getch() != '^') {
if (opd.empty()) {
b += 1;
} else {
b += opd.top();
opd.pop();
}
}
ungetch();
break;
}
}
if (!opd.empty()) {
c = opd.top();
opd.pop();
}
double delta, x1, x2;
delta = pow(b, 2) - 4*a*c;
if (delta >= 0) {
delta = sqrt(delta);
x1 = (-b-delta) / (2*a);
x2 = (-b+delta) / (2*a);
if (x1 > x2)
swap(x1, x2);
printf("%.2lf %.2lf\n", x1, x2);
} else {
printf("No Solution\n");
}
}
return 0;
}
发表于 2019-02-14 14:46:39
回复(0)
def getNum(equation,digit): #得到该次幂前面的系数
index = equation.find(digit)
if index >= 0:
temp = '#' #用一个非数字符代替特殊的1被隐藏
index -= 1
while index >= 0 and(equation[index].isdigit() or equation[index] == '-'):
temp = equation[index] + temp
index -= 1
if temp == '#':
return 1
elif temp == '-#':
return -1
else:
return int(temp[:-1])
return 0
while True:
try:
equation = input().split('=')
a = getNum(equation[0],'x^2')-getNum(equation[1],'x^2')
for i in range(2):
if equation[i].find('x^2') != -1:
equation[i] = equation[i][equation[i].find('x^2')+3:]
b = getNum(equation[0],'x')-getNum(equation[1],'x')
for i in range(2):
if equation[i].find('x') != -1:
equation[i] = equation[i][equation[i].find('x')+1:]
c = 0 #只要得到方程ax^2+bx+c=0中的a，b，c，之后就很好算
if equation[0]:
c = int(equation[0])
if equation[1]:
c -= int(equation[1])
if b*b-4*a*c < 0:
print('No Solution')
else:
x1 = (-b+(b*b-4*a*c)**0.5)/(2*a)
x2 = (-b-(b*b-4*a*c)**0.5)/(2*a)
if x1 > x2:
x1,x2 = x2,x1
print('%.2f %.2f'%(x1,x2))
except Exception:
break
编辑于 2018-10-14 14:10:53
回复(0)
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<stack>
#include<cmath>
using namespace std;
int cba[3],epos;
char str[100];
void Read(int &co,int &eo,int &i){
if(str[i]==0) {eo=-1; return ;}
int flag=0;
if(str[i]=='=') {++i;}
if(str[i]=='-') {++i;++flag;}
if(str[i]=='+') ++i;
if(i>epos) ++flag;
if(str[i]=='x'){
co=1;
if(str[i+1]=='^'){i+=2; eo=str[i]-'0'; ++i;}
else {i++; eo=1;}
}
else{
co=0;
while(str[i]>='0'&&str[i]<='9')
{co=co*10+str[i]-'0'; ++i;}
if(str[i]=='x'){
if(str[i+1]=='^'){i+=2; eo=str[i]-'0';++i;}
else {i++; eo=1;}
}
else {eo=0;++i;}
}
if(flag%2){
co=0-co;
}
}
int main(){
while(scanf("%s",str)==1){
for(int i=0;;++i){
if(str[i]==0)break;
if(str[i]=='=') {epos=i; break;}
}
int co,eo,i=0;
memset(cba,0,sizeof(cba));
while(1){
Read(co,eo,i);
if(eo==-1) break;
cba[eo]+=co;
}
int a=cba[2],b=cba[1],c=cba[0];
// printf("a:%d,b:%d,c:%d\n",a,b,c);
double delta=b*b-4*a*c;
if(delta<0) printf("No Solution\n");
else {
double x1,x2;
x1=(0-b-sqrt(delta))/(2*a);
x2=(0-b+sqrt(delta))/(2*a);
if(x1>x2) swap(x1,x2);
printf("%.2f %.2f\n",x1,x2);
}
}
return 0;
}
发表于 2018-05-06 19:06:36
回复(0)
//我觉得我的代码可以往上爬一爬
#include<stdio.h>
#include<math.h>#include<string.h>
#include<algorithm>
using namespace std;
int main(){
char str[100];
int eqPos;// =位置
while(scanf("%s",str)!=EOF){
int a=0,b=0,c=0; //a为二次项系数，b为一次项系数，c为常数
for(int i=0;i<strlen(str);i++){ //找到等号的位置，后面比较用
if(str[i]=='='){
eqPos=i;
}
}
for(int i=0;i<strlen(str);i++){ //一次遍历即可
if(str[i]=='x'){ //不是常数项
if(str[i+1]=='^'){ //是二次的，则求其系数
if(i==0)a=a+1; //在开头的x^2，系数为1
else{
int t=0,k=1;
int site=i-1;
while(site>=0&&str[site]>='0'&&str[site]<='9'){ //往前倒，算系数
t=(str[site]-'0')*k+t;
k*=10;
site--;
}
if(site==i-1)t=1;//qianm单独的x^2，系数为1
if(i>eqPos){
if(str[site]!='-')a-=t;
else a+=t;
}
else{
if(str[site]!='-')a+=t;
else a-=t;
}
}
i+=2;// x^2 后移两位继续识别
}else{ //一次项x
if(i==0)b=b+1;//x在开头，系数为1
else{
int t=0,k=1;
int site=i-1;
while(site>=0&&str[site]>='0'&&str[site]<='9'){//往前倒，求出系数
t=(str[site]-'0')*k+t;
k*=10;
site--;
}
if(site==i-1)t=1;//单独的x 系数为1
if(i>eqPos){//等号右边
if(str[site]!='-')b-=t;//符号为+
else b+=t;
}
else{//等号左边
if(str[site]!='-')b+=t;//符号为+
else b-=t;
}
}
}
}else if(str[i]>='0'&&str[i]<='9'){ //暂且认为是常数项
int site=i;
int t=0;
while(site<strlen(str)&&str[site]>='0'&&str[site]<='9'){//不管是不是常数，先把数算出来备用
t=str[site]-'0'+t*10;
site++;
}
if(str[site]!='x'){ //如果遇到x，就不是常数了
if(i>eqPos){
if(str[i-1]!='-')c-=t;
else c+=t;
}
else{
if(str[i-1]!='-')c+=t;
else c-=t;
}
i=site;
}
}else{}
}
//printf("a=%d,b=%d,c=%d\n",a,b,c);
double ans1,ans2;
if(b*b-4*a*c>=0){
ans1=(double)(-b-sqrt(b*b-4*a*c))/(2*a);
ans2=(double)(-b+sqrt(b*b-4*a*c))/(2*a);
if(ans1>ans2){ //小的在前
swap(ans1,ans2);
}
printf("%.2f %.2f\n",ans1,ans2);
}else{
printf("No Solution\n");
}
}
return 0;
}
发表于 2019-03-23 16:57:40
回复(0)
可以用状态机来做
#include <stdio.h>
#include <math.h>
#define WAIT 0
#define MEET_NUM 1
#define MEET_X 2
int is_num(char c){
int tmp=(int)c;
return (tmp >=(int)'0' && tmp<=(int)'9');
}
int main(){
char tmp;
int side=1;
int a=0,b=0,c=0;
int state=WAIT;
int current_num=0;
int sign=1;
while(scanf("%c",&tmp)!=EOF){
if(tmp=='-'){
sign=-1;
}
if(tmp=='=') side=-1;
switch(state){
case WAIT:if(tmp=='x'){
state=MEET_X;
current_num=sign*side;
break;
}
if(is_num(tmp)){
current_num=((int)tmp-(int)'0')*sign*side;
state=MEET_NUM;
break;
}
break;
case MEET_NUM:if(tmp=='x'){
state=MEET_X;
break;
}
else{
if(is_num(tmp)){
int t=(int)tmp-(int)'0';
current_num=current_num*10+t*side*sign;
break;
}
else{
c+=current_num;
sign=1;
state=WAIT;
break;
}
}
case MEET_X:if(tmp=='^'){
a+=current_num;
sign=1;
scanf("%c",&tmp);
state=WAIT;
break;
}
else{
b+=current_num;
sign=1;
state=WAIT;
break;
}
}
}
if(state==MEET_X) b+=current_num;
if(state==MEET_NUM) c+=current_num;
float b2_4ac=(float)(b*b-4*a*c);
if(b2_4ac<0){
printf("No Solution\n");
}
else{
float result1=((float)b*(-1.0)-fabs((sqrt(b2_4ac))))/(2*a);
float result2=((float)b*(-1.0)+fabs((sqrt(b2_4ac))))/(2*a);
if(result2<result1){
float temp=result2;
result2=result1;
result1=temp;
}
printf("%.2f %.2f\n",result1,result2);
}
}
#include <math.h>
#define WAIT 0
#define MEET_NUM 1
#define MEET_X 2
int is_num(char c){
int tmp=(int)c;
return (tmp >=(int)'0' && tmp<=(int)'9');
}
int main(){
char tmp;
int side=1;
int a=0,b=0,c=0;
int state=WAIT;
int current_num=0;
int sign=1;
while(scanf("%c",&tmp)!=EOF){
if(tmp=='-'){
sign=-1;
}
if(tmp=='=') side=-1;
switch(state){
case WAIT:if(tmp=='x'){
state=MEET_X;
current_num=sign*side;
break;
}
if(is_num(tmp)){
current_num=((int)tmp-(int)'0')*sign*side;
state=MEET_NUM;
break;
}
break;
case MEET_NUM:if(tmp=='x'){
state=MEET_X;
break;
}
else{
if(is_num(tmp)){
int t=(int)tmp-(int)'0';
current_num=current_num*10+t*side*sign;
break;
}
else{
c+=current_num;
sign=1;
state=WAIT;
break;
}
}
case MEET_X:if(tmp=='^'){
a+=current_num;
sign=1;
scanf("%c",&tmp);
state=WAIT;
break;
}
else{
b+=current_num;
sign=1;
state=WAIT;
break;
}
}
}
if(state==MEET_X) b+=current_num;
if(state==MEET_NUM) c+=current_num;
float b2_4ac=(float)(b*b-4*a*c);
if(b2_4ac<0){
printf("No Solution\n");
}
else{
float result1=((float)b*(-1.0)-fabs((sqrt(b2_4ac))))/(2*a);
float result2=((float)b*(-1.0)+fabs((sqrt(b2_4ac))))/(2*a);
if(result2<result1){
float temp=result2;
result2=result1;
result1=temp;
}
printf("%.2f %.2f\n",result1,result2);
}
}
发表于 2018-08-28 13:51:14
回复(0)
# 正则表达式解法
import re
import mathfrom itertools import chain
def get_coef(regex, lexp, rexp):
coef = 0
pos = re.finditer(regex, lexp)
neg = re.finditer(regex, rexp)
for match in chain(pos, neg):
num = match.group(1)
num = 1 if len(num) == 0 else int(num)
coef = coef + num if "=" in match.string else coef - num
return coef
if __name__ == '__main__':
lexp, rexp = input().split("=")
lexp += "="
coef2 = get_coef(r"((?:\+|-|)\d+|)x\^2", lexp, rexp);
coef1 = get_coef(r"((?:\+|-|)\d+|)x(?!\^2)", lexp, rexp);
coef0 = get_coef(r"(?<![\d^])((?:\+|-|)\d+)(?![\dx])", lexp, rexp);
# print("%d %d %d" %(coef2, coef1, coef0))
try:
delta = math.sqrt(coef1**2 - 4 * coef2 * coef0)
except:
print("No Solution")
exit(0)
rt1 = (-coef1 - delta) / (2 * coef2)
rt2 = (-coef1 + delta) / (2 * coef2)
if rt1 > rt2:
rt1, rt2 = rt2, rt1
print("%.2f %.2f" % (rt1, rt2))
编辑于 2018-07-01 11:05:03
回复(0)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int getNum(char* s,int *m,int i){
int c=0,num=0;
while(s[i]>='0'&&s[i]<='9'){
num=num*10+s[i]-'0';
++i;
++c;
}
*m=c;
return num;
}
double root(int p){
double root=1.0;
while(root*root-p>=1e-9||root*root-p<=-1e-9)
root=(root+p/root)/2;
return root;
}
void solution(int a,int b,int c){
int p=b*b-4*a*c;
double s1,s2,ro=0.0;
if(p<0)
printf("No Solution\n");
else{
ro=root(p);
a*=2;
b=-b;
s1=(b+ro)/a;
s2=(b-ro)/a;
printf("%.2f %.2f\n",s2,s1);
}
}
int main(){
char f[100];
scanf("%s",f);
int temp=1,*mov,i=0, a=0,b=0,c=0,flag=1,sig=1;
while(f[i]){
temp=0;
if(f[i]>='0'&&f[i]<='9'){
temp=getNum(f,mov,i)*flag*sig;//record the parameter
i+=*mov;
if(f[i]!='x')
c+=temp;
else if(f[i+1]=='^'&&f[i+2]=='2'){
a+=temp;
i+=3;
}
else{
b+=temp;
++i;
}
}
else if(f[i]=='+'||f[i]=='-'||f[i]=='='){
switch (f[i]) {
case '+':
++i;
sig=1;
break;
case '-':
sig=-1;
++i;
break;
case '=':
flag=-1;
sig=1;
++i;
break;
}
}
else if(f[i]=='x'){
if(f[i+1]=='^'&&f[i+2]=='2'){
a+=sig*flag;
i+=3;
}
else{
b+=sig*flag;
i+=1;
}
}
}
solution(a, b, c);
}//为什么不对啊，，，我自己的机器上可以的，。提示信息：运行错误:请检查是否存在数组、列表等越界非法访问，内存非法访问等情况
发表于 2017-01-11 23:55:53
回复(0)
虽然被该题的标签给难住了（说是二分，但实际上利用求根公式就可以了。因此，该题的难点在于如何处理将给定的字符串的相关系数提取出来。
发表于 2022-02-12 15:35:43
回复(0)
#include
#include
#include
#include
#include
#include
using namespace std;
void cal(int a2, int a1, int a0) {
int delt = a1 * a1 - 4 * a2 * a0;
double d = 0.0;
double num1, num2;
if (delt < 0) {
cout << "No Solution";
} else {
d = sqrt(delt);
num1 = (0 - a1 - d) / 2;
num2 = (0 - a1 + d) / 2;
printf("%.2f %.2f", num1, num2);
}
}
int main() {
string str;
cin >> str;
int a2 = 0;
int a1 = 0;
int a0 = 0;
queue q;
char op = '+';
int t = 0;
int i;
for ( i = 0; i < str.size() && str[i] != '='; i++) {
if (str[i] >= '0' && str[i] <= '9') {
q.push(str[i] - '0');
} else if (str[i] == 'x') {
i++;
if (str[i] == '^') {
if (q.empty())a2 += 1;
while (!q.empty()) {
t = t * 10 + q.front();
q.pop();
}
if (op == '+')a2 += t;
else a2 -= t;
t = 0;
i++;
} else {
if (q.empty())a1 += 1;
while (!q.empty()) {
t = t * 10 + q.front();
q.pop();
}
if (op == '+')a1 += t;
else a1 -= t;
t = 0;
i--;
}
} else {
while (!q.empty()) {
t = t * 10 + q.front();
q.pop();
}
if (op == '+')a0 += t;
else a0 -= t;
t = 0;
op = str[i];
}
}
if (str[i] == '=') {
//处理前面可能遗留的常数项
while (!q.empty()) {
t = t * 10 + q.front();
q.pop();
}
if (op == '+')a0 += t;
else a0 -= t;
t = 0;
i++;
op = '+';
for ( ; i < str.size() ; i++) {
if (str[i] >= '0' && str[i] <= '9') {
q.push(str[i] - '0');
} else if (str[i] == 'x') {
i++;
if (str[i] == '^') {
if (q.empty())a2 += 1;
while (!q.empty()) {
t = t * 10 + q.front();
q.pop();
}
if (op == '+')a2 -= t;
else a2 += t;
t = 0;
i++;
} else {
if (q.empty())a1 += 1;
while (!q.empty()) {
t = t * 10 + q.front();
q.pop();
}
if (op == '+')a1 -= t;
else a1 += t;
t = 0;
i--;
}
} else {
while (!q.empty()) {
t = t * 10 + q.front();
q.pop();
}
if (op == '+')a0 -= t;
else a0 += t;
t = 0;
op = str[i];
}
}
}
while (!q.empty()) {
t = t * 10 + q.front();
q.pop();
}
if (op == '+')a0 -= t;
else a0 += t;
t = 0;
cal(a2, a1, a0);
}
// 64 位输出请用 printf("%lld")
发表于 2025-02-28 00:54:05
回复(0)
parse部分代码：
int co[3];
int getLevel(string& t, int& p) {
if (p + 1 >= t.length() || t[p + 1] != '^') {
return 1;
} else {
p += 2;
return t[p] - '0';
}
}
void parse(string& t) {
co[0] = co[1] = co[2] = 0;
int right = 1;
int nagtive = 1;
int now = 0;
for (int i = 0; i < t.length(); i++) {
if (isdigit(t[i]))
now = now * 10 + t[i] - '0';
else {
if (t[i] == 'x') {
if (now == 0) now = 1;
co[getLevel(t, i)] += right * nagtive * now;
nagtive = 1;
now = 0;
} else {
if (now != 0) {
co[0] += right * nagtive * now;
nagtive = 1;
now = 0;
}
if (t[i] == '-')
nagtive = -1;
else if (t[i] == '=')
right = -1;
}
}
}
if (now != 0)
co[0] += right * nagtive * now;
}
编辑于 2024-03-22 18:28:39
回复(0)
#include <cmath>
#include <cstdlib>
#include <iomanip>
#include <iostream>
#include <string>
using namespace std;
struct Polynomial { //（一元）二次多项式
int a, b, c; //系数（a：二次项系数，b：一次项系数，c：常数项）
Polynomial(string str) { //构造函数
int position = str.find("x^2");
if (position != -1) { //无二次项
a = atoi(str.substr(0, position).c_str());
str = str.substr(position + 3);
if (str[0] == '+') { //一次项系数为正
str = str.substr(1); //吃掉加号
}
}
position = str.find("x");
if (position != -1) { //无一次项
b = atoi(str.substr(0, position).c_str());
str = str.substr(position + 1);
if (str[0] == '+') { //常数项为正
str = str.substr(1); //吃掉加号
}
}
c = atoi(str.c_str());
}
};
struct Equation { //（一元）二次方程
Polynomial* left; //方程左式
Polynomial* right; //方程右式
Equation(string str) { //构造函数
int position = str.find("=");
left = new Polynomial(str.substr(0, position));
right = new Polynomial(str.substr(position + 1));
}
void solve() { //方程求解
int a = left->a - right->a;
int b = left->b - right->b;
int c = left->c - right->c;
int delta = b * b - 4 * a * c;
if (delta < 0) {
cout << "No Solution" << endl;
} else {
double x1 = (-b * 1.0 - sqrt(delta)) / 2 * a;
double x2 = (-b * 1.0 + sqrt(delta)) / 2 * a;
cout << setiosflags(ios_base::fixed) << setprecision(2) << x1 << " "
<< setiosflags(ios_base::fixed) << setprecision(2) << x2 << endl;
}
}
};
int main() {
string str;
while (cin >> str) {
Equation(str).solve();
}
return 0;
}
编辑于 2024-03-01 15:22:56
回复(0)
#include <bits/stdc++.h>
using namespace std;
void js(string s1, int& xs1, int& xs2, int& cs) {
s1 = '+' + s1 + "+0+";
for (int i = 0; i < s1.size(); i++)
if (s1[i] == 'x' && s1[i + 1] != '^') {
if (i == 0 || s1[i - 1] == '+') xs2++;
if (i > 0 && s1[i - 1] == '-') xs2--;
if (i > 0) {
int j = i - 1;
int u = 0, k = 1;
while (j >= 0 && s1[j] >= '0' && s1[j] <= '9') {
u = u + (s1[j] - '0') * k;
k *= 10;
j--;
}
if (j < 0 || s1[j] == '+') xs2 += u;
else if (s1[j] == '-') xs2 -= u;
}
} else if (s1[i] == 'x' && s1[i + 1] == '^') {
if (i == 0 || s1[i - 1] == '+') xs1++;
if (i > 0 && s1[i - 1] == '-') xs1--;
if (i > 0) {
int j = i - 1;
int u = 0, k = 1;
while (j >= 0 && s1[j] >= '0' && s1[j] <= '9') {
u = u + (s1[j] - '0') * k;
k *= 10;
j--;
}
if (j < 0 || s1[j] == '+') xs1 += u;
else if (s1[j] == '-') xs1 -= u;
}
}
int i = 0;
while (i < s1.size())
if (s1[i] >= '0' && s1[i] <= '9' && s1[i - 1] != '^') {
int j = i;
int u = 0;
while (s1[j] >= '0' && s1[j] <= '9') {
u = u * 10 + s1[j] - '0';
j++;
}
if (s1[i - 1] == '-') u = -u;
if (s1[j] != 'x') cs += u;
i = j;
} else i++;
}
int main() {
string s;
cin >> s;
string s0 = "", s2 = "";
int n = s.size();
for (int i = 0; i < s.size(); i++)
if (s[i] == '=') {
for (int j = 0; j < i; j++)
s0 = s0+(char)s[j];
for (int j = i+1; j < s.size(); j++)
s2 = s2+(char)s[j];
break;
}
int r = 0, p = 0, g = 0;
js(s0, r, p, g);
int aa = 0, bb = 0, cc = 0;
js(s2, aa, bb, cc);
int a = r - aa;
int b = p - bb;
int c = g - cc;
int delta = b*b-4*a*c;
double x1 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);
double x2 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);
int a1 = (int)x1;
int a2 = (int)x2;
int aa1 = abs((int)((x1-a1)*1000));
int aa2 = abs((int)((x1-a1)*1000));
if (aa1 > 0 && aa1 %10 >= 5) aa1 +=10;
if (aa2 > 0 && aa2 %10 >= 5) aa2 +=10;
if (aa1 < 0 && (-aa1)% 10 >= 5) aa1 -= 10;
if (aa2 < 0 && (-aa2)% 10 >= 5) aa2 -= 10;
cout << a1 << '.' << abs(aa1)/10 << ' ';
cout << a2 << '.' << abs(aa2)/10 << '\n';
return 0;
}
发表于 2023-03-20 15:49:39
回复(0)
//将所有变量和常量都移到等式左边，得到aX^2+bX+c=0,在用公式做
#include "stdio.h"
#include "math.h"
#include "string"
#include "iostream"
#include "algorithm"
using namespace std;
string str;
int calculate(int pos){//计算x^2,x前的数字(num1,num2)的位数
int count = 0;
for (int i = pos-1; i >= 0; --i) {
if (str[i] >= '0' && str[i] <= '9'){
++count;
continue;
} else
break;
}
++count;
return count;
}
int calculate2(int pos){//计算普通数字num3的位数
int count = 0;
for (int i = pos; str[i] >= '0' && str[i] <= '9'; ++i) {
++count;
}
return count;
}
int main(){
;int sum1=0,sum2=0,sum3=0;//sum1为x^2系数,sum2为x系数，sum3为常数(都移到方程左侧)
while (getline(cin,str)){
/*
* 对str初始化，对x,x^2变成1x，1x^2，开头的再变成+/-1x，+/-x^2，相当于规格化，便于后续处理
*/
for (int i = 0; i < str.size(); ++i) {//加1
if (str[i] == 'x'){
if (str[i-1] == '+' || str[i-1] == '-' || str[i-1] == '=' || i==0){
str.insert(i,"1");
continue;
}
}
if (str[i] > '0' && str[i] <= '9'){
if (str[i-1] == '=' || i == 0){
str.insert(i,"+");
continue;
}
}
}
if (str[1] == 'x')//加"+"
str.insert(0,"+");
int pos_equal = str.find("=");
if (str[pos_equal+2] == 'x')
str.insert(pos_equal+1,"+");
while (str.find("x^2")!=string::npos){//算出sum1 ->a
int pos = str.find("x^2");
int count = calculate(pos);
if (pos < pos_equal){
sum1 += stoi(str.substr(pos-count,count));
str.erase(pos-count,count+3);
} else{
sum1 -= stoi(str.substr(pos-count,count));
str.erase(pos-count,count+3);
}
pos_equal = str.find("=");
}
while (str.find("x") != string::npos){//算sum2 ->b
int pos = str.find("x");
int count = calculate(pos);
if (pos < pos_equal){
sum2 += stoi(str.substr(pos-count,count));
str.erase(pos-count,count+1);
} else{
sum2 -= stoi(str.substr(pos-count,count));
str.erase(pos-count,count+1);
}
pos_equal = str.find("=");
}
for (int i = 0; i < str.size(); ++i) {//算num3
if (str[i] > '0' && str[i] <= '9'){
int count = calculate2(i);
if (i < pos_equal){
sum3 += stoi(str.substr(i-1,count+1));
str.erase(i-1,count+1);
} else{
sum3 -= stoi(str.substr(i-1,count+1));
str.erase(i-1,count+1);
}
pos_equal = str.find("=");
}
}
double temp = sum2*sum2*1.0-4*sum1*sum3;
if (temp<0)
printf("No Solution\n");
else{
double x = (-sum2 - sqrt(temp))/(2*sum1);
double y = (-sum2 + sqrt(temp))/(2*sum1);
if (x > y)
swap(x,y);
printf("%.2llf %.2llf\n",x,y);
}
}
}
发表于 2023-03-13 19:01:17
回复(0)
#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <cstring>
using namespace std;
int main(){
string str;
int a1 = 0, b1 = 0, c1 = 0; //左边系数
int a2 = 0, b2 = 0, c2 = 0; //右边系数
while(cin >> str){
int len = str.size();
bool right = false; //判断等式左右
bool Symbol = true; //+号为true，-号为false
int temp = 0;
for(int i = 0; i < len; i++){
if(!right){ //等式左边
if(str[i] == '='){
if(temp != 0){
c1 = temp;
if(!Symbol) c1 = -c1;
temp = 0;
}
right = true;
Symbol = true;
}
else if(isdigit(str[i])) temp = 10 * temp + str[i] - '0';
else if(str[i] == '+' || str[i] == '-') Symbol = str[i] == '+';
else if(str[i] == 'x'){
if(temp == 0) temp = 1;
if(i + 1 < len && str[i + 1] == '^'){
a1 = temp;
if(!Symbol) a1 = -a1;
i += 2;
}
if(i + 1 < len && str[i + 1] != '^'){
b1 = temp;
if(!Symbol) b1 = -b1;
}
temp = 0;
}
}else{ //等式右边
if(isdigit(str[i])) temp = 10 * temp + str[i] - '0';
else if(str[i] == '+' || str[i] == '-') Symbol = str[i] == '+';
else if(str[i] == 'x'){
if(temp == 0) temp = 1;
if(i + 1 < len && str[i + 1] == '^'){
a2 = temp;
if(!Symbol) a2 = -a2;
i += 2;
}
if(i + 1 < len && str[i + 1] != '^'){
b2 = temp;
if(!Symbol) b2 = -b2;
}
temp = 0;
}
}
}
if(temp != 0){
c2 = temp;
if(!Symbol) c2 = -c2;
}
double x1, x2, temp1, temp2;
a1 -= a2;
b1 -= b2;
c1 -= c2;
temp1 = b1 * b1 - 4 * a1 * c1;
if(temp1 < 0) printf("No Solution\n");
else{
temp2 = sqrt(temp1);
x1 = (-b1 + temp2) / 2 / a1;
x2 = (-b1 - temp2) / 2 / a1;
if(x1 < x2) printf("%.2lf %.2lf\n", x1, x2);
else printf("%.2lf %.2lf\n", x2, x1);
}
}
return 0;
}
发表于 2022-03-05 14:02:13
回复(0)
#include"iostream"
#include"math.h"
#include"iomanip"
using namespace std;
struct equation{
int a, b, c;
equation(int aa, int bb, int cc):a(aa), b(bb), c(cc){}
equation(){}
};
equation getCoff(string ipt, int equ){// 自左向右扫描串
equation e(0, 0, 0); // 根据数字，+,-,x,^调整临时系数对应的符号，次数，值
int flag= 1, temp= 0, power= 0; // 每次扫描后查看是否进行系数的更新
for(int i= 0; i< ipt.length(); i++){
if(ipt[i]== '+'|| ipt[i]== '-'){flag= ipt[i]== '+'? 1: -1;}
else if(isdigit(ipt[i])){temp= temp* 10+ ipt[i]- '0';}
else if(ipt[i]== 'x'){power++;}
else if(ipt[i]== '^'){power= power- 1+ ipt[i+ 1]- '0'; i++;}
if(i+ 1== ipt.length()|| ipt[i+ 1]== '+'|| ipt[i+ 1]== '-'){
if(power== 0){e.c+= temp* flag* equ;}
else if(power== 1){temp= (temp== 0? 1:temp); e.b+= temp* flag* equ;}// 防止一次项系数为1的情况
else{temp= (temp== 0? 1:temp); e.a+= temp* flag* equ;}// 防止二次型系数为1的情况
flag= 1, temp= 0, power= 0;// 初始化临时系数的符号，值，次数
}
}
return e;
}
void killRoot(equation e){// 求根
float deta= (pow(e.b, 2)- 4*e.a*e.c);// b^2- 4ac
if(deta>= 0){
float r1= (-e.b -sqrt(deta))/(2* e.a* 1.0), r2= (-e.b +sqrt(deta))/ (2* e.a* 1.0);
cout<< fixed<< setprecision(2)<< min(r1, r2)<< " "<< max(r1, r2)<< endl;
}else{
cout<< "No Solution"<< endl;
}
}
int main(){
string ipt;
while(getline(cin, ipt)){
int idx= ipt.find('='); // 确定等号
string left= ipt.substr(0, idx);// 分解等式
string right= ipt.substr(idx+ 1);
equation e1= getCoff(left, 1);// 获取等式左侧的二次式系数
equation e2= getCoff(right, -1);// 获取等式右侧的二次式系数
equation e(e1.a+ e2.a, e1.b+ e2.b, e1.c+ e2.c);// 完整的二次式系数
killRoot(e);// 求解
}
}
发表于 2021-05-03 10:56:38
回复(0)
//以等号为边界，找到两边的系数再进行计算
//找系数过程：找到数字，看后面是x、x^2还是其他
//注意测试案例全部先输出较小的根
#include<iostream>
#include<string>
#include<vector>
#include<cmath>
#include<iomanip>
using namespace std;
vector<int> count(int begin, int end, string str){
vector<int> num(3, 0);
int tmp=0;
int op=0;
for(int i=begin; i<=end; i++){
if(str[i]>='0' && str[i]<='9'){
//判断数字的符号
if(i==begin || str[i-1]=='+') op=1;
else if(str[i-1]=='-') op=-1;
//数字可能有多位，连续读入
tmp=str[i]-'0';
while(str[i+1]>='0' && str[i+1]<='9'){
i++;
tmp=tmp*10+str[i]-'0';
}
tmp*=op;//乘上符号
if(i==end) num[2]+=tmp;//最后一位是数字，肯定是常数项
//后面一位是x
else if(str[i+1]=='x'){
i++;//跳过这个x
if(i+1==end) num[1]+=tmp;//最后一位是x，肯定是一次项
else if(str[i+1]=='^') {//二次项
num[0]+=tmp;
i+=2;//跳过^x
}
else num[1]+=tmp;//数字后面不是^2，则是一次项
}
else num[2]+=tmp;//数字后面不是x，则是常数项
}
}
return num;
}
int main(){
string str;
vector<int> left(3), right(3);
double a, b, c, tmp;
double x1, x2, deta;
while(cin>>str){
//对x^2，x这种隐藏系数1的进行补充
for(int i=0; i<str.size(); i++){
if(str[i]=='x'){
if(i==0){
str.insert(str.begin()+i, '1');
i++;
}
else if(str[i-1]<'0' || str[i-1]>'9'){
str.insert(str.begin()+i, '1');
i++;
}
}
}
int index=str.find('=');
left=count(0, index-1, str);//等号左边系数
right=count(index+1, str.size()-1, str);//等号右边系数
a=left[0]-right[0];
b=left[1]-right[1];
c=left[2]-right[2];
//计算根
deta=b*b-4*a*c;
if(deta<0) cout<<"No Solution"<<endl;
else{
x1=(-sqrt(deta)-b)/(2*a);
x2=(sqrt(deta)-b)/(2*a);
if(x1>x2) swap(x1, x2);//为了AC，先输出小的根
cout<<fixed<<setprecision(2)<<x1<<' '<<x2<<endl;
}
}
return 0;
}
发表于 2021-03-11 23:32:43
回复(0)
Java解法
1. 首先将-都替换为+- （这样可以保留负数）；
2. 以=为分割符分割为左右半边；
3. 左右半边以+为分割符，分别分成若干个未知数例如5x^2
4. 提取左右半边系数，然后移项后得a，b，c
5. 利用(-b+-(b^2-4ac))/2a进行求解
（其中要注意 double保留两位小数，以及判断是否有解）
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
/**
*
* 求解 5x^2+x=3x+4
* 求根公式 (-b+—(b^2-4ac))/2a
* @author csidez
*
*/
public class Main {
//解析未知数的系数
void solveArray(String[] array, Map<String, Integer> map) {
for (int i=0; i<array.length; i++) {
if (array[i].contains("x^2")) {
String num = array[i].substring(0, array[i].indexOf("x^2"));
map.merge("x^2", num.equals("")?1:Integer.parseInt(num), (oldV, newV)->oldV+newV);
} else if (array[i].contains("x")) {
String num = array[i].substring(0, array[i].indexOf("x"));
map.merge("x", num.equals("")?1:Integer.parseInt(num), (oldV, newV)->oldV+newV);
} else {
map.merge("C", array[i].equals("")?0:Integer.parseInt(array[i]), (oldV, newV)->oldV+newV);
}
}
}
public static void main(String args[]) {
Scanner scanner = new Scanner(System.in);
Main fc = new Main();
String line[] = scanner.nextLine().replace("-", "+-").split("\\=");
String left[] = line[0].split("\\+");
String right[] = line[1].split("\\+");
Map<String, Integer> lmap = new HashMap<String, Integer>();
Map<String, Integer> rmap = new HashMap<String, Integer>();
fc.solveArray(left, lmap);
fc.solveArray(right, rmap);
double a = lmap.getOrDefault("x^2", 0) - rmap.getOrDefault("x^2", 0);
double b = lmap.getOrDefault("x", 0) - rmap.getOrDefault("x", 0);
double c = lmap.getOrDefault("C", 0) - rmap.getOrDefault("C", 0);
double k = Math.sqrt(b*b-4*a*c);
if (b*b-4*a*c < 0) {
System.out.println("No Solution");
} else {
System.out.println(String.format("%.2f", (-b+k)/(2*a)) + " " + String.format("%.2f", (-b-k)/(2*a)));
}
}
}
发表于 2021-03-03 16:35:42
回复(0)
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