Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 1815 Accepted
Submission(s): 929
1 #include <iostream>
2 #include <stdio.h>
3 #include <cmath>
4 using namespace std;
5
6 int main()
7 {
8 int n;
9 while(scanf("%d",&n)!=EOF){
10 int sum = 0;
11 for(int i=0;i<=sqrt(n);i++){
12 int j = n-i*i;
13 double t = sqrt(j);
14 if(t-int(t)==0){ //开方成功
15 if(i!=0 && j!=0)
16 sum+=4;
17 else if(i==0 || j==0)
18 sum+=2;
19 }
20 }
21 printf("%d\n",sum);
22 }
23 return 0;
24 }
Freecode : www.cnblogs.com/yym2013
hdu 3835:R(N)(水题,数学题),布布扣,bubuko.com
原文:http://www.cnblogs.com/yym2013/p/3596791.html