Problem Description
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).
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)题解:暴力枚举一下,要注意if(i>j)break;这一句,否则会重复计算WA掉。
#include <cstdio> #include <iostream> #include <string> #include <cstring> #include <stack> #include <queue> #include <algorithm> #include <cmath> #include <map> using namespace std; //#define LOCAL int main() { #ifdef LOCAL freopen("in.txt", "r", stdin); #endif // LOCAL //Start int n; while(cin>>n) { int ans=0; for(int i=0; i<sqrt(n); i++) { double j=sqrt(n-i*i); if(i>j)break; if((int)j==j) { if(i==0||j==0||i==j)ans+=4; else ans+=8; } } printf("%d ",ans); } return 0; }