HDU 3835R(N)(加点思维的暴力枚举)

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 DescriptionWe 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). 
InputNo more than 100 test cases. Each case contains only one integer N(N<=10^9). 
OutputFor each N, print R(N) in one line. 
Sample Input

26102565

 
Sample Output

4081216HintFor 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) 

 
Source2011 Multi-University Training Contest 1 - Host by HNU 
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。
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