R(N)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1637 Accepted Submission(s): 853
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 16HintFor 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)
Source
题目大意：题目意思很好懂，找有多少组a,b可以满足a^2+b^2==n.n最大是10^9,肯定不可以直接暴力枚举。那就枚举a>b>=0，这样的情况。然后每一种情况会*4。a,b;a,-b;-a,b;-a,-b;
如果b=0的话，5,0;-5,0;0,-5,0,5;也是*4；详见代码。
如果b=0的话，5,0;-5,0;0,-5,0,5;也是*4；详见代码。
题目地址：R(N)
AC代码：
#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
int n,i,a,b,ma,res;
while(~scanf("%d",&n))
{
ma=sqrt(double(n)); //最大的a,a>b>=0
res=0;
for(a=1;a<=ma;a++)
{
b=sqrt(double(n-a*a));
if(a*a+b*b==n)
res+=4; //a,b;a,-b;-a,b;-a,-b;
}
cout<<res<<endl;
}
return 0;
}
//46MS 280K