R(N)
HUST - 1544Description
We know that some positive integer x can be expressed as x=A^2+B^2(A,B are integers). Take x=10 for example, 10=(-3)^2+1^2.
We define R(N) (N is positive) to be the total number of variable presentation of N. So R(1)=4, which consists of 1=1^2+0^2, 1=(-1)^2+0^2, 1=0^2+1^2, 1=0^2+(-1)^2.Given N, you are to calculate R(N).
Input
No more than 100 test cases. Each case contains only one integer N(N<=10^9).
Output
For each N, print R(N) in one line.
Sample Input
2 6 10 25 65
Sample Output
4 0 8 12 16
Hint
For the fourth test case, (A,B) can be (0,5), (0,-5), (5,0), (-5,0), (3,4), (3,-4), (-3,4), (-3,-4), (4,3) , (4,-3), (-4,3), (-4,-3)
题意:给出一个数n,问有多少种方式使得n=a*a+b*b
解题思路:构造
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
#include <functional>
using namespace std;
#define LL long long
const int INF=0x3f3f3f3f;
int main()
{
int n,sum,k;
while(~scanf("%d",&n))
{
sum=0;
if(n==0) printf("1\n");
else
{
k=sqrt(1.0*n/2);
for(int i=0;i<=k;i++)
{
int j=sqrt(n-i*i);
if(i*i+j*j==n)
{
if(i==0) sum+=4;
else if(i==j) sum+=4;
else sum+=8;
}
}
printf("%d\n",sum);
}
}
return 0;
}