hdu 3835 简单概率dp

R(N)

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 3   Accepted Submission(s) : 1

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

26102565

 

Sample Output

4081216
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)

 

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int n;int main(){    while(scanf("%d",&n)!=EOF)    {        if(!n)        {            printf("1\n");            continue;        }        int i,j,ans=0,ave=(int)sqrt(double(n/2.0));        for(i=ave; i>=0; i--)        {             j=(int)sqrt(1.0*(n-i*i));                if(i*i+j*j==n)                {                    if(i==0||j==0||i==j) ans+=4;                    else ans+=8;                }        }        printf("%d\n",ans);    }    return  0;}/*26102565*/

Source

2011 Multi-University Training Contest 1 - Host by HNU