HDU 3835 R(N)

R(N)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2761    Accepted Submission(s): 1420


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).
 
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)
 
Source
 
Recommend
xubiao
暴力题,注意不要超时就行了,注意sqrt的数一定得是int类型的。
include <iostream>
#include <math.h>
#include <stdio.h>
using namespace std;
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
        {
            cout<<1<<endl;
            continue;
        }
        int c=sqrt(n/2);
        int ans=0;
        int i,j;
        for(i=0;i<=c;i++)
        {
            j=sqrt(n-i*i);
            if(i*i+j*j==n)
            {
                if(i==0||j==0||i==j)
                ans=ans+4;
                else
                ans=ans+8;
            }
        }
        printf("%d
",ans);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/Ritchie/p/5432349.html