HDU 3835R(N)(加点思维的暴力枚举)

R(N) p>Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

p>Total Submission(s): 1637    Accepted Submission(s): 853

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p>Problem Description

p>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, 

p>10=(-3)^2+1^2.

p>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).

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p>Input

p>No more than 100 test cases. Each case contains only one integer N(N<=10^9).

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p>Output

p>For each N, print R(N) in one line.

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p>Sample Input

p>2

p>6

p>10

p>25

p>65

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p>Sample Output

p>4

p>0

p>8

p>12

p>16

p>Hint

p>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)

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p>Source

p>2011 Multi-University Training Contest 1 - Host by HNU

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p>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。

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p>题目地址：R(N)

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p>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