<cf> Funky Numbers

最新推荐文章于 2022-03-28 11:02:27 发布

原创最新推荐文章于 2022-03-28 11:02:27 发布 · 786 阅读

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文章标签：

#numbers #integer #input

本文介绍了一个算法问题：如何判断一个给定的整数是否可以表示为两个三角形数（即三角数）的和。文章提供了一种有效的解决方案，并附带了完整的代码实现，该方案仅使用一个循环即可完成任务。

As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as $\text{[math]}$ , where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 10⁹).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Sample test(s)

input
256

output
YES

input
512

output
NO

Note

In the first sample number $\text{[math]}$ .

In the second sample number 512 can not be represented as a sum of two triangular numbers.

一重for loop解决，不用两重

AC Code

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
    int n,sr;
    bool flag;
    while(cin>>n)
    {
        flag=0;
        sr=pow(2*n-3,0.5);
        for(int i=1;i<=sr;i++)
        {
            int k=pow(2*n-i*(i+1),0.5);
            //do not neglect "k>0"
            if(k && k*(k+1)+i*(i+1)==2*n)
            {
                flag=1;
                break;
            }
        }
        if(flag) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}