小喵们很喜欢把自己装进容器里的（例如碗），但是要是碗的周长比喵的身长还短，它们就进不去了。
现在告诉你它们的身长，和碗的半径，请判断一下能否到碗里去。
[编程题]快到碗里来
long double防止溢出。
#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<functional>
#include <map>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <exception>
using namespace std;
int main(int argc, char** argv)
{
long double n,r;
while (cin >> n >> r)
{
if (2 * r * 3.14 >= n) cout << "Yes" << endl;
else cout << "No" << endl;
}
return 0;
}
发表于 2017-07-08 15:44:17
回复(0)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
Double n = sc.nextDouble();
Double r = sc.nextDouble();
if(n < 2 * 3.14 * r) System.out.println("Yes");
else System.out.println("No");
}
}
}
发表于 2016-10-15 23:14:26
回复(1)
#include <stdio.h>
#include <stdlib.h>
#define P 3.14
#define CIRCUMFERENCE(r) (2*(P)*(r))
int is_completion(const int height, const int radius)
{
int res = 0;
res = ((CIRCUMFERENCE(radius) < height) ? (1 : 0));
return res;
}
int main(int argc, char const *argv[])
{
/* code */
int res;
int height, radius;
printf("please input the height of cat, and input the radius:");
scanf("%d,%d", &height, &radius);
res = is_completion(&height, &radius);
if (res)
{
/* code */
printf("NO\n");
}
else{
printf("YES\n");
}
return 0;
}
编辑于 2015-08-31 17:04:54
回复(0)
正确的做法是使用大数乘法，明显n和r的精度已经超出了内置类型的表示范围了。
#include <iostream>
#include <string>
using namespace std;
string times628(string & str)
{
string ret(str.length() + 3, 0);
string str628 = "628";
int strL = str.length();
int retL = ret.length();
int c = 0;
for (int k = 2; k >= 0; --k)
for (int i = strL - 1, j = retL + k - 3; i >= 0; --i, --j)
ret[j] += (str[i] - '0') * (str628[k] - '0');
for (int i = retL; i >= 0; --i)
{
ret[i] += c;
c = ret[i] / 10;
ret[i] = ret[i] % 10 + '0';
}
int index = 0;
while (ret[index] == '0') ++index;
return string(ret.begin() + index, ret.end() - 2);
}
int compare(string & a, string & b)
{
if (a.length() != b.length())
return (a.length() > b.length() ? 1 : -1);
else
{
for (int i = 0; i < a.length(); ++i)
if (a[i] == b[i]) continue;
else return (a[i] > b[i] ? 1 : -1);
return 0;
}
}
int main()
{
string n, r;
while (cin >> n >> r)
{
r = times628(r);
if (compare(n, r) == 1) cout << "No" << endl;
else cout << "Yes" << endl;
}
return 0;
}
发表于 2015-12-14 19:27:16
回复(1)
还是我实诚,手写了一个链表实现乘法
import java.util.LinkedList;
import java.util.Scanner;
public class Main {
//小猫咪钻罐子
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()){
//猫
String s1 = scanner.next();
//罐子
String s2 = scanner.next();
LinkedList<Integer> list1 = new LinkedList<>();
LinkedList<Integer> list2 = new LinkedList<>();
for (int i = 0; i < s1.length(); i++) {
list1.add(s1.charAt(i)-'0');
}
for (int i = 0; i < s2.length(); i++) {
list2.add(s2.charAt(i)-'0');
}
int size = list2.size()-1;//对应最后一个元素得下标
for (int i = size; i >= 0; i--) {
//将size下标得值*2
list2.set(i,list2.get(i)*2);
}
//再全部进数
for (int i = size; i >= 0 ; i--) {
if(list2.get(i) >= 10){
//进位,将前面得值+1
if(i == 0){
//如果是头插就插入1
list2.addFirst(1);
list2.set( 1 ,list2.get(1) % 10);
}else {
//如果不是头插就修改值
list2.set( i-1 ,list2.get(i-1)+1);
//进位后将当前值%10
list2.set( i ,list2.get(i) % 10);
}
}
}
// System.out.println(list2);
size = list2.size()-1;//重新更新下size的值
//Πd,此时d处理了,就差*Π了
//*314
LinkedList<Integer> list = new LinkedList<>();
int[]pai = {3,1,4};
int n1 = 0;//记录相加
int n2 = 0;//记录相乘
int z = 0;//一直记录倒数第二个位置
for (int i = 2; i >= 0 ; i--) {
n1++;
n2 = n1;
for (int j = size; j >= 0 ; j--) {
z = list.size()-n2;
int p = list2.get(j);
//说明此时没有对应的值
if(z == -1){
list.addFirst(p * pai[i]);
z++;
}else {
int tmp = list.get(z);
list.set(z,p * pai[i]+tmp);
}
if(list.get(z) >= 10){
//如果为0,说明需要头插
if(z == 0){
list.addFirst(list.get(z)/10);
z++;
}else {
//如果不为0,就给前面的值加上进的位数
int tmp = list.get(z-1);
list.set(z-1,list.get(z) / 10+tmp);
}
list.set(z,list.get(z) % 10);
}
n2++;
}
}
// System.out.println(list);
//做个标记,记录这个数组从哪里开始是小数
int len = list.size()-2;
//从后往前数,去二位0
int i = 0;
while (i < 2){
//获取后数两位
//如果是0就删除
if(list.get(list.size()-1) == 0){
list.pollLast();
}
i++;
}
size = 0;//记录次数
//类似于 12 111
if(list1.size() > len){
System.out.println("No");
}else if(list1.size() < len){
System.out.println("Yes");
}else {
//两个一样长时
while (list1.size() != 0 && list.size() !=0){
//如果猫的身长小于碗的周长
if(list1.peekFirst() < list.peekFirst()){
System.out.println("Yes");
break;
}else if(list1.peekFirst() > list.peekFirst()){
System.out.println("No");
break;
}else {
list1.pollFirst();
list.pollFirst();
}
size++;
}
if(list1.size() == 0 && (list.get(0)!=0 || list.get(1)!=0))
System.out.println("Yes");
}
}
}
}
发表于 2023-04-18 16:45:48
回复(0)
#include <iostream>
#include<vector>
#include<algorithm>
using namespace std;
void getcomp(string a, string b) {
if (a.length() < b.length()) {
cout << "Yes" << endl;
} else if (a.length() == b.length()) {
auto lt = a.begin();
auto lb = b.begin();
while (lt != a.end()) {
if (*lt > *lb) {
cout << "No" << endl;
break;
}
if(*lt<*lb){
cout << "Yes" << endl;
break;
}
lt++;
lb++;
if(lt==a.end()&&lb==b.end()){
cout<<"Yes"<<endl;
}
}
} else
cout << "No" << endl;
}
string getsum(string a, string b) {
auto la = a.begin();
auto lb = b.begin();
string str;
int adv = 0;
while (la != a.end() && lb != b.end()) {
int tmp = (*la - '0') + (*lb - '0') + adv;
str += (tmp % 10 + '0');
adv = tmp / 10;
la++;
lb++;
}
while (la != a.end()) {
int tmp = (*la - '0') + adv;
str += (tmp % 10 + '0');
adv = tmp / 10;
la++;
}
while (lb != b.end()) {
int tmp = (*lb - '0') + adv;
str += (tmp % 10 + '0');
adv = tmp / 10;
lb++;
}
return str;//倒序
}
void constr(string a, string b) {
a += "00";
reverse(b.begin(), b.end());//小在前
int ar[3] = { 8, 2, 6 };
vector<string>str(3);
for (int i = 0; i < 3; i++) {
auto la = b.begin();
int adv = 0;
while (la != b.end()) {
int tmp = ar[i] * (*la - '0') + adv;
str[i] += (tmp % 10 + '0');
adv = tmp / 10;
la++;//小数在高位 倒序
}
if (adv != 0)
str[i] += (adv + '0');
//reverse(str[i].begin(), str[i].end());
int j = i;
while (j) {
str[i].insert(str[i].begin(), '0');
j--;
}
}
string abs;
abs = getsum(str[0], str[1]);
abs = getsum(abs, str[2]);
reverse(abs.begin(), abs.end());
getcomp(a, abs);
}
int main() {
string st1, st2;
while (cin >> st1 >> st2 ) {
constr(st1, st2);
}
return 0;
} 唯一要注意的地方是 getcomp这个比较函数
从高位到低位 只要有a[i]>b[i ]就是No；只要有a[i]< b[i]就是Yes; 两判断条件缺一不可
发表于 2023-08-08 09:38:38
回复(0)
import java.math.BigDecimal;
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
//方法1：使用double来做
//while(in.hasNext()){
double n = in.nextDouble();//double 范围 -1.7 E—308~1.7E-308 是1.7*10的308次方
double r = in.nextDouble();
if(n > (2*3.14*r)){
System.out.println("No");
}else{
System.out.println("Yes");
}
}//
//方法2：BigDecimal来做，BigDecimal可以用来算大数（任意精度的十进制数（浮点数））
while(in.hasNext()){
String n =in.next();
String r =in.next();
BigDecimal a =new BigDecimal(n);//a是身长
BigDecimal b = new BigDecimal(r);//b是半径
BigDecimal c = new BigDecimal(6.28);//3.14*2
b=b.multiply(c);//周长：b*c
if(a.compareTo(b)>0){//若a>b 返回1
System.out.println("No");
}else{
System.out.println("Yes");
}
}
}
}
发表于 2023-04-19 16:39:29
回复(0)
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()){
BigInteger[] integers = new BigInteger[2];
integers[0] = scanner.nextBigInteger();//喵的身长
integers[1] = scanner.nextBigInteger();//碗的半径
if ((integers[1].doubleValue() * 2 * 3.14) < integers[0].doubleValue()){
System.out.println("No");
}else {
System.out.println("Yes");
}
}
}
}
发表于 2023-03-30 20:05:03
回复(0)
#include <iostream>
using namespace std;
int main()
{
double n,r;
while(cin >> n >> r)
{
if(2*3.14*r > n){
cout << "Yes" << endl;
}else{
cout << "No" << endl;
}
}
return 0;
}
using namespace std;
int main()
{
double n,r;
while(cin >> n >> r)
{
if(2*3.14*r > n){
cout << "Yes" << endl;
}else{
cout << "No" << endl;
}
}
return 0;
}
import java.util.*;
import java.io.*;
import java.math.BigDecimal;
public class Main{
public static void main(String[] args){
Scanner scan = new Scanner(System.in);
while(scan.hasNext()){
BigDecimal n = scan.nextBigDecimal();
BigDecimal r = scan.nextBigDecimal();
BigDecimal len = new BigDecimal("6.28").multiply(r);
System.out.println(n.compareTo(len) == 1?"No":"Yes");
}
}
}
发表于 2023-02-09 16:53:34
回复(0)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (in.hasNext()) {
double n = in.nextDouble();
double r = in.nextDouble();
if(n >= 2 * 3.14 * r){
System.out.println("No");
}else{
System.out.println("Yes");
}
}
}
}
发表于 2022-05-04 19:51:54
回复(0)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()) {
Double cat = scanner.nextDouble();
Double perimeter = scanner.nextDouble();
if(cat < 2 * 3.14 * perimeter){
System.out.println("Yes");
}
else {
System.out.println("No");
}
}
}
}
发表于 2022-03-21 21:37:55
回复(0)
import java.util.*;
import java.math.*;
public class Main{
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
while(scanner.hasNext()){
String str1 = scanner.next();
String str2 = scanner.next();
BigDecimal bigInteger1 = new BigDecimal(str1);
BigDecimal bigInteger2 = new BigDecimal(str2);
BigDecimal bigInteger3 = new BigDecimal("3.14");
bigInteger2 = bigInteger2.multiply(bigInteger3.multiply(new BigDecimal("2")));
int num = bigInteger1.compareTo(bigInteger2);
if(num == -1){
System.out.println("Yes");
}else if(num == 0){
System.out.println("Yes");
}else if(num == 1){
System.out.println("No");
}
}
}
}
发表于 2021-05-19 19:57:52
回复(0)
问题信息
难度：
28条回答
28收藏
10623浏览
通过挑战的用户
查看代码-
2023-03-12 21:32:50 -
2023-03-12 20:43:15
-
2023-03-11 09:41:07
-
2023-03-09 14:40:49
-
2023-03-09 10:25:29
-
扫描二维码，关注牛客网
- 意见反馈
-
下载牛客APP，随时随地刷题
扫一扫，把题目装进口袋
- 求职之前，先上牛客
-
扫描二维码，进入QQ群
扫描二维码，关注牛客公众号
- 公司地址：北京市朝阳区北苑路北美国际商务中心K2座一层-北京牛客科技有限公司
- 联系方式：010-60728802 投诉举报电话：010-57596212（朝阳人力社保局）
- 牛客科技© All rights reserved admin@nowcoder.com
- 京ICP备14055008号-4 增值电信业务经营许可证 营业执照 人力资源服务许可证
-
京公网安备
11010502036488号
