[编程题]Primary Arithmetic
- 热度指数：2828 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 64M，其他语言128M
- 算法知识视频讲解
Children are taught to add multi-digit numbers from right-to-left one digit at a time. Many find the "carry" operation - in which a 1 is carried from one digit position to be added to the next - to be a significant challenge. Your job is to count the number of carry operations for each of a set of addition problems so that educators may assess their difficulty.
输入描述:
Each line of input contains two unsigned integers less than 10 digits.
输出描述:
For each line of input except the last you should compute and print the number of carry operations that would result from adding the two numbers, in the format shown below.
示例1
输入
123 456 555 555 123 594 0 0
输出
NO carry operation. 3 carry operations. 1 carry operation.
#include<iostream>
(720)#include<string>
using namespace std;
string a,b;
int main()
{
while(cin>>a>>b&&a!="0"&&b!="0")
{
int indexa=a.length()-1;
int indexb=b.length()-1;
int carry=0;
int count=0;
while(indexa>=0||indexb>=0)
{
if(a[indexa--]-'0'+b[indexb--]-'0'+carry>9)
{
carry=1;
count++;
}
else
carry=0;
}
if(count==0)
cout<<"NO carry operation."<<endl;
else if(count==1)
cout<<count<<" carry operation."<<endl;
else
cout<<count<<" carry operations."<<endl;
}
return 0;
}
发表于 2020-04-27 15:13:26
回复(2)
//看了大家的题解，感觉大家方法都好高级，自己想的简单了点也是过了哈哈哈
//注意0进位，1进位，多进位的输出格式^_^
#include<iostream>
using namespace std;
int main()
{
int a,b;
while(cin>>a>>b)
{
bool flag = false; //判断是否从右往左有进位
if(a==0&&b==0) break;
int tmp1=0,tmp2=0,sum=0,cnt = 0;
while(a>0||b>0)
{
tmp1 = a%10;
tmp2 = b%10;
cnt = tmp1+tmp2;
if(flag==true) cnt =tmp1+tmp2+1; //若有进位
flag = false; //进位清空
if(cnt>=10)
{
flag = true;
sum++;
}
a/=10;
b/=10;
}
if(sum == 0) cout<<"NO carry operation."<<endl;
else if(sum == 1) cout<<sum<<" carry operation."<<endl;
else cout<<sum<<" carry operations."<<endl;
}
return 0;
}
发表于 2020-04-13 12:14:35
回复(0)
#include<stdio.h>
#include<string.h>
int main()
{
char a[100],b[100];
int c[100]={0},d[100]={0},i,num;
scanf("%s%s",a,b);
for(i=strlen(a)-1,num=0;i>=0;i--)//1.字符数组转整数数组第一位放个位容易计算
c[num++]=a[i]-'0';
for(i=strlen(b)-1,num=0;i>=0;i--)
d[num++]=b[i]-'0';
int max= strlen(a)>strlen(b)?strlen(a):strlen(b);
for(i=0,num=0;;i<max;i++)//2.解决进位问题
if(c[i]+d[i]>9)//需要进位
{ num++;c[i+1]+=1; }//上一位加1
if(num==0) printf("NO carry operation.\n");
else if(num==1) printf("1 carry operation.\n");
else printf("%d carry operations.\n",num);
}
编辑于 2020-04-02 21:43:00
回复(0)
一招大数，横扫相加类的问题
#include <bits/stdc++.h>
using namespace std;
int main(){
string s1,s2;
int a[110]= {0},b[110]= {0};
while(cin>>s1>>s2){
int flag=0;
for(int i=0; i<s1.size(); i++) {
a[i]=s1[s1.size()-i-1]-'0';
}
int len=max(s1.size(),s2.size())+1;
for(int i=0; i<len; i++) {
if(i<s2.size()){
b[i]=s2[s2.size()-i-1]-'0';
}
a[i]+=b[i];
if(a[i]>=10){
a[i+1]++;
a[i]%=10;
flag++;
}
}
if(flag==0) cout<< "NO carry operation."<<endl;
else if(flag==1)cout<< "1 carry operation."<<endl;
else cout<<flag<<" carry operations."<<endl;
}
return 0;
}
发表于 2020-03-21 17:51:49
回复(0)
看来我写的还不算是太麻烦，就注意一点，就是输出数字的时候也要以x carry operations的形式输出，开始就是没有以这种形式，所以有一次没有通过
#include<iostream>
#include<cstring>
using namespace std;
int main(){
char a[10],b[10];
while(cin>>a>>b){
int lenA=strlen(a);
int lenB=strlen(b);
int tempB=lenB-1;
int count=0;
int temp=0;
for(int i=lenA-1;i>=0;i--){
if((a[i]-'0'+b[tempB--]-'0'+temp)>=10){
count++;
temp=1;
}
else temp=0;
}
if(count==1) cout<<count<<" carry operation."<<endl;
else if(count>1) cout<<count<<" carry operations."<<endl;
else cout<<"NO carry operation."<<endl;
}
}
发表于 2019-02-09 11:21:03
回复(1)
while True:
try:
a,b = input().split()
if len(a) > len(b):
a,b = b,a #保证a是比较短的
carry = count = 0
while a: #每次相加最后一位和进位，并在循环结尾去掉最后一位
if int(a[-1]) + int(b[-1]) + carry >= 10:
carry = 1
count += 1
else:
carry = 0
a = a[:-1]
b = b[:-1]
while b:
if int(b[-1]) + carry >= 10:
carry = 1
count += 1
else:
carry = 0
b = b[:-1]
if count == 0:
print("NO carry operation.")
elif count == 1:
print("1 carry operation.") #注意输出细节
else:
print('%d carry operations.' % count)
except Exception:
break
编辑于 2018-10-01 22:20:08
回复(0)
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int main(){
char number1[12], number2[12];
while (cin >> number1 >> number2){
int len1 = strlen(number1);
int len2 = strlen(number2);
int carry[12];
int count = 0;
memset(carry, 0, sizeof(carry));
for (int i = len1-1, j = len2-1; j >= 0 || i >= 0; j--, i--){
int number = number1[i] + number2[j] - '0' - '0' + carry[max(i, j) + 1];
if (number>=10){
carry[max(i, j)] = number / 10;
count++;
}
}
if (len1>len2){
for (int i = (len1 - len2) - 1; i >= 0; i--){
if (number1[i] + carry[i + 1] - '0'>=10){
carry[i] = ((number1[i] + carry[i + 1]-'0') / 10);
count++;
}
}
}
else{
for (int i = (len2 - len1) - 1; i >= 0; i--){
if (number2[i] + carry[i + 1]-'0'>=10){
carry[i] = ((number2[i] + carry[i + 1]-'0') / 10);
count++;
}
}
}
if (count > 1)
cout << count << " carry operations."<<endl;
else if (count == 0){
cout <<"No carry operations." << endl;
}else
cout << count << " carry operation." << endl;
}
}
发表于 2018-09-15 15:35:28
回复(0)
__author__ = 'Yaicky'
def calCarry(num1, num2, carry):
carry = (num1 + num2 + carry) / 10
return carry
while True:
a, b = map(int, raw_input().strip().split())
if a==0 and b==0:
break
carry = count = 0
while a%10 or b%10 or a or b: #例842079407 799043235 不加上or a和or b会提前结束while
carry = calCarry(a%10, b%10, carry)
if carry:
count += 1
a /= 10
b /= 10
if count == 0:
print 'NO carry operation.'
elif count == 1:
print '1 carry operation.'
else:
print '%d carry operations.' % (count)
发表于 2016-06-27 11:55:39
回复(0)
def fun(a):
res = []
while a >0:
t = a % 10
a = a / 10
res.append(t)
return res
def add(a, b):
a = fun(a)
b = fun(b)
i, j = 0, 0
c = 0
while i < len(a) and j < len(b):
if a[i] + b[j] >= 10:
c+=1
i+=1
j+=1
return c
while True:
try:
a, b = map(int, input().split())
if a == 0 and b == 0:
break
res = add(a, b)
if res == 0:
print("NO carry operation.")
elif res == 1:
print("%d carry operation."%(res))
else:
print("%d carry operations."%(res))
except:
break
发表于 2024-03-21 22:51:44
回复(0)
#include <iostream>
#include <string>
using namespace std;
string Reverse(string a){
string res;
for(int i=a.size()-1; i>=0; --i){
res.push_back(a[i]);
}
return res;
}
int Add(string a, string b){
int temp = 0, cnt = 0;
a = Reverse(a);
b = Reverse(b);
for(int i=0; i<a.size() || i<b.size(); ++i){
if(i<a.size()) temp+=a[i]-'0';
if(i<b.size()) temp+=b[i]-'0';
temp /= 10;
if(temp) cnt++;
}
return cnt;
}
int main(){
string a, b;
while(cin >> a >> b){
if(a == "0" && b == "0") break;
int cnt = Add(a, b);
if(cnt>1) cout << cnt << " carry operations." << endl;
else if(cnt>0) cout << cnt << " carry operation." << endl;
else cout << "NO carry operation." << endl;
}
}
编辑于 2024-03-11 21:35:27
回复(0)
#include <iostream>
#include <istream>
#include <string>
using namespace std;
const int MAXN = 10;
struct MyInteger { //自定义整数类，用数组表示整数
int* digit; //数组表示数值各位
int length; //数组长度（位数）
MyInteger(int x);
MyInteger operator=(string str);
};
MyInteger::MyInteger(int x) {
digit = new int[MAXN]();
*this = to_string(x);
}
MyInteger MyInteger::operator=(string str) {
length = str.length();
for (int i = 0; i < length; i++) {
digit[i] = str[length - i - 1] - '0';
}
return *this;
}
//统计加法进位次数
int countCarry(const MyInteger& a, const MyInteger& b) {
int count = 0, carry = 0;
for (int i = 0; i <= a.length && i <= b.length && i < 10; i++) {
carry = (a.digit[i] + b.digit[i] + carry) / 10;
count += carry;
}
return count;
}
int main() {
int a, b;
while (cin >> a >> b && a && b) {
int result = countCarry(MyInteger(a), MyInteger(b));
result == 0 ? cout << "NO" : cout << result;
cout << " carry operation";
if (result > 1) {
cout << "s";
}
cout << "." << endl;
}
return 0;
}
编辑于 2024-03-01 00:12:52
回复(0)
//大意为每个测试用例给两个整数，计算这两个整数相加的过程中产生的总进位数。
//多个测试用例，最后一行的0，0不用算
/*
*感觉用字符串做更方便些
*/
#include "stdio.h"
#include "string"
#include "algorithm"
using namespace std;
int main(){
char num1[11],num2[12];
while(scanf("%s %s",num1,num2)!=EOF){
string str1 = num1,str2 = num2;
if (str1 == "0" && str2 == "0")
break;
reverse(str1.begin(),str1.end());//低字节存储地位,这样的话个位在左，不逆转的话个位在右，
//从右向左做，思路都一样，没有什么区别
reverse(str2.begin(),str2.end());
int sum = 0,carry = 0;//记录进位次数的,记录当前是否进位
int length = max(str1.size(),str2.size());
while (str1.size() < length)
str1 += "0";
while (str2.size() < length)
str2 += "0";
for (int i = 0; i < length; ++i) {
if(str1[i]-'0'+str2[i]-'0'+carry >= 10){
carry = 1;
++sum;
}
else
carry = 0;
}
if (sum == 0)
printf("NO carry operation.\n");
else if(sum == 1)
printf("1 carry operation.\n");
else
printf("%d carry operations.\n",sum);
}
}
发表于 2023-03-10 16:10:46
回复(0)
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <malloc.h>
int main()
{
char num1[10];
char num2[10];
while(scanf("%s %s",num1,num2)!=EOF)
{
if(strcmp(num1,"0")==0&&strcmp(num2,"0")==0)
{
break;
}
int len1=strlen(num1);
int len2=strlen(num2);
for(int i=0;i<len1/2;i++)
{
int t=num1[i];
num1[i]=num1[len1-1-i];
num1[len1-1-i]=t;
}
for(int i=0;i<len2/2;i++)
{
int t=num2[i];
num2[i]=num2[len2-1-i];
num2[len2-1-i]=t;
}
/*
for(int i=0;i<len1;i++)
{
printf("%d ",num1[i]-'0');
}
printf("\n");
for(int i=0;i<len2;i++)
{
printf("%d ",num2[i]-'0');
}
*/
int count=0;
int c=0;
for(int i=0;i<len1&&i<len2;i++)
{
int a=num1[i]+num2[i]+c-'0'-'0';
if(a>=10)
{
//printf("o %d %d",num1[i]-'0',num2[i]-'0');
count++;
c=1;
}else
{
c=0;
}
}
if(len1<=len2)
{
for(int i=len1;i<len2;i++)
{
int a=num2[i]-'0'+c;
if(a>=10)
{
printf("p ");
count++;
c=1;
}else
{
c=0;
}
}
}
if(len1>len2)
{
for(int i=len2;i<len1;i++)
{
int a=num1[i]-'0'+c;
if(a>=10)
{
printf("q ");
count++;
c=1;
}else
{
c=0;
}
}
}
if(count==0)
{
printf("NO carry operation.\n");
}else if(count ==1)
{
printf("%d carry operation.\n",count);
}else
{
printf("%d carry operations.\n",count);
}
}
}
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <malloc.h>
int main()
{
char num1[10];
char num2[10];
while(scanf("%s %s",num1,num2)!=EOF)
{
if(strcmp(num1,"0")==0&&strcmp(num2,"0")==0)
{
break;
}
int len1=strlen(num1);
int len2=strlen(num2);
for(int i=0;i<len1/2;i++)
{
int t=num1[i];
num1[i]=num1[len1-1-i];
num1[len1-1-i]=t;
}
for(int i=0;i<len2/2;i++)
{
int t=num2[i];
num2[i]=num2[len2-1-i];
num2[len2-1-i]=t;
}
/*
for(int i=0;i<len1;i++)
{
printf("%d ",num1[i]-'0');
}
printf("\n");
for(int i=0;i<len2;i++)
{
printf("%d ",num2[i]-'0');
}
*/
int count=0;
int c=0;
for(int i=0;i<len1&&i<len2;i++)
{
int a=num1[i]+num2[i]+c-'0'-'0';
if(a>=10)
{
//printf("o %d %d",num1[i]-'0',num2[i]-'0');
count++;
c=1;
}else
{
c=0;
}
}
if(len1<=len2)
{
for(int i=len1;i<len2;i++)
{
int a=num2[i]-'0'+c;
if(a>=10)
{
printf("p ");
count++;
c=1;
}else
{
c=0;
}
}
}
if(len1>len2)
{
for(int i=len2;i<len1;i++)
{
int a=num1[i]-'0'+c;
if(a>=10)
{
printf("q ");
count++;
c=1;
}else
{
c=0;
}
}
}
if(count==0)
{
printf("NO carry operation.\n");
}else if(count ==1)
{
printf("%d carry operation.\n",count);
}else
{
printf("%d carry operations.\n",count);
}
}
}
发表于 2022-02-26 18:40:37
回复(0)
大数改编不止十位也行
#include <iostream>
using namespace std;
int countCarry(string a, string b) {
int cnt = 0, carry = 0;
if (a.length() < b.length()) {
swap(a, b);
}
int i, j;
for (i = a.length() - 1, j = b.length() - 1; j >= 0; i--, j--) {
carry = (carry + a[i] - '0' + b[j] - '0') / 10;
if (carry) {
cnt++;
}
}
while (i >= 0) {
carry = (carry + a[i] - '0') / 10;
if (carry) {
cnt++;
}
i--;
}
return cnt;
}
int main() {
string a, b;
int cnt;
while (cin >> a >> b) {
if (a == "0" && b == "0") {
continue;
}
cnt = countCarry(a, b);
if (cnt == 0) {
cout << "NO carry operation." << endl;
}
else if (cnt == 1) {
cout << "1 carry operation." << endl;
}
else {
cout << cnt << " carry operations." << endl;
}
}
return 0;
}
发表于 2021-03-14 15:54:17
回复(0)
给咱玩文字游戏呢？？大小写， 单复数？？
#include<iostream>
#include<cstdio>
using namespace std;
const int maxn = 15;
int change(long a, int num[]) {
int i = 0;
while(a > 0) {
num[i++] = a % 10;
a /= 10;
}
return i;
}
int main() {
long a, b;
while(cin >> a >> b) {
if(a == 0 && b == 0) break;
int num1[maxn] = {0}, num2[maxn] = {0};
int lena = change(a, num1);
int lenb = change(b, num2);
int ans = 0, carry = 0;
int i;
for(i = 0; i < lena || i < lenb; ++i) {
int temp = num1[i] + num2[i] + carry;
if(temp >= 10) {
carry = temp / 10;
ans++;
} else {
carry = 0;
}
}
if(ans == 0) printf("NO carry operation.\n");
else if(ans == 1) printf("1 carry operation.\n");
else printf("%d carry operations.\n", ans);
}
return 0;
}
发表于 2021-02-24 20:55:51
回复(0)
字符串从0到len-1依次存放的是数字的高位到低位，与加法思维模式相反，
反序存储到数组中，使数组0到len-1依次存放的是数字的低位到高位，
然后模拟加法运算（记得memset清空数组，避免上一组样例影响下一组）
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;
int main(){
string a, b;
int aa[15], bb[15];
while(cin >> a >> b){
if(a=="0" && b=="0")
break;
memset(aa, 0, sizeof(aa));
memset(bb, 0, sizeof(bb));
int alen = a.length(), blen = b.length();
for(int i = alen-1; i >= 0; i--)
aa[alen-1-i] = a[i]-'0';
for(int i = blen-1; i >= 0; i--)
bb[blen-1-i] = b[i]-'0';
int cnt = 0, c = 0, tmp = 0;
for(int i = 0; i<alen || i<blen; i++){
tmp = aa[i]+bb[i]+c;
c = tmp/10;
if(c)
cnt++;
}
if(cnt == 0)
printf("NO carry operation.\n");
else if(cnt == 1)
printf("1 carry operation.\n");
else
printf("%d carry operations.\n", cnt);
}
return 0;
}
编辑于 2020-09-18 14:59:07
回复(0)
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
void reverse(char str[]){
int len=strlen(str);
for(int i=0;i<len/2;++i){
swap(str[i],str[len-i-1]);
}
}
int main(){
char str1[12],str2[12];
while(scanf("%s %s",str1,str2)!=EOF){
int len=max(strlen(str1),strlen(str2));
reverse(str1);
reverse(str2);
int count=0;
int carry=0;
for(int i=0;i<len;++i){
carry=(str1[i]-'0'+str2[i]-'0'+carry)/10;
if(carry==1)
count++;
}
if(count==0){
printf("NO carry operation.\n");
}else if(count==1){
printf("%d carry operation.\n",count);
}
else{
printf("%d carry operations.\n",count);
}
}
return 0;
}
编辑于 2020-04-20 19:28:09
回复(0)
#include<bits/stdc++.h>
using namespace std;
int main(){
string s1,s2;
while(cin>>s1>>s2){
if(s1=="0"&&s2=="0")break;
int len1=s1.length(),len2=s2.length();
int i=len1-1,j=len2-1,carry=0,c=0;
while(i>=0||j>=0||carry>0){
int n1 = i>=0?s1[i]-'0':0;
int n2 = j>=0?s2[j]-'0':0;
if(carry+n1+n2>=10)c++;
carry=(carry+n1+n2)/10;
i--;
j--;
}
if(c==0){
cout<<"NO carry operation."<<endl;
}else if(c==1){
cout<<"1 carry operation."<<endl;
}else{
cout<<c<<" carry operations."<<endl;
}
}
return 0;
}
发表于 2020-04-09 11:47:10
回复(0)
#include<stdio.h>
(737)#include<string.h>
int main()
{
char a[11],b[11];
int cnt;
while(scanf("%s %s",&a, &b)!=EOF)
{
if(a == 0 && b == 0) break;
int len_a = strlen(a), len_b = strlen(b);
cnt = 0;
int k = 0, t, i = len_a-1, j = len_b-1;
while( i > -1 || j > -1)
{
t = k;
if(i > -1) t+=a[i--]-'0';
if(j > -1) t+=b[j--]-'0';
k = t/10;
if(k == 1) cnt++;
}
if(cnt == 0) printf("NO carry operation.\n");
else if(cnt == 1) printf("%d carry operation.\n",cnt);
else printf("%d carry operations.\n",cnt);
}
return 0;
}
发表于 2020-03-31 17:13:51
回复(0)
#include <bits/stdc++.h>
using namespace std;
int cmp(string x, string y){
int a[1001];
int b[1001];
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
int count=0;
int x1=x.size();
int y1=y.size();
int m=0;
for(int i=1001-x1;i<1001;++i){
a[i]=x[m++]-'0';
}
m=0;
for(int i=1001-y1;i<1001;++i){
b[i]=y[m++]-'0';
}
if(x1<y1){
for(int i=1000;i>=1001-x1;--i){
b[i]=a[i]+b[i];
}
for(int i=1000;i>0;--i){
if(b[i]>9){
count++;
b[i-1]++;
b[i]-=10;
}
}
}
else{
for(int i=1000;i>=1001-y1;--i){
a[i]=a[i]+b[i];
}
for(int i=1000;i>0;--i){
if(a[i]>9){
count++;
a[i-1]++;
a[i]-=10;
}
}
}
return count;
}
int main(){
string x,y;
while(cin>>x>>y){
if(x=="0"&&y=="0")
break;
int count = cmp(x,y);
if(count==0)
cout<<"NO carry operation."<<endl;
else if(count==1)
cout<<count<<" carry operation."<<endl;
else
cout<<count<<" carry operations."<<endl;
}
return 0;
}
发表于 2020-03-18 17:24:46
回复(0)
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