hdu 3835(R(N)) 一道简单方程折射出的编程思维

R(N)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1458    Accepted Submission(s): 762

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)

 The hint in the problem is very importtant,make sure read is carefully!

 

 

 

Problem analysis:

Give you one integer N,then you need to calculate the occurrence number of the equation "i*i+j*j=N";

本题虽小，但代表了一类问题的解法。也能展现一个人的面对问题时的思维：

本题有多种遍历方法：其中i可以从0到sqrt(1.0*N)遍历；也可以从0到sqrt(1.0*N/2)遍历；

第二种遍历的方法更高效，故采用第二种遍历方法；

 

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int N;
    while(cin>>N)
    {
        int t=(int)sqrt(N/2.0);
        int i,j;
        int count=0;
        for(i=0;i<=t;i++)
        {
            j=(int)sqrt(1.0*N-i*i);
            if(i*i+j*j==N)
            {
                if(i==0||j==0||i==j)
                 count+=4;
                else
                    count+=8;
            }

        }
        cout<<count<<endl;
    }
    return 0;
}