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  • HDU 3835 R(N)(枚举)

    题目链接

    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)
     
    题解:暴力枚举一下,要注意if(i>j)break;这一句,否则会重复计算WA掉。
    #include <cstdio>
    #include <iostream>
    #include <string>
    #include <cstring>
    #include <stack>
    #include <queue>
    #include <algorithm>
    #include <cmath>
    #include <map>
    using namespace std;
    //#define LOCAL
    
    int main()
    {
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
    #endif // LOCAL
        //Start
        int n;
        while(cin>>n)
        {
            int ans=0;
            for(int i=0; i<sqrt(n); i++)
            {
                double j=sqrt(n-i*i);
                if(i>j)break;
                if((int)j==j)
                {
                    if(i==0||j==0||i==j)ans+=4;
                    else ans+=8;
                }
            }
            printf("%d
    ",ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/gpsx/p/5160623.html
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