Funky Numbers

最新推荐文章于 2023-04-01 18:00:32 发布

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最新推荐文章于 2023-04-01 18:00:32 发布

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分类专栏：数论

本文链接：https://blog.youkuaiyun.com/qq_40941611/article/details/81269975

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本文介绍了一种算法，用于判断一个给定的整数是否可以表示为两个三角形数之和。通过预先计算一系列三角形数，并使用二分查找来高效地寻找可能的配对，实现了这一目标。

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As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , 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).

Examples

Input

Output

YES

Input

Output

NO

Note

In the first sample number .

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

这个题我先打了一个表，然后遍历第一个，再用二分找第二个。

#include <iostream>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <cmath>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
typedef long long ll;
const ll maxn=1e9+10;
ll f[1000000];
int main()
{   ll s,d;
s=d=1;
int flag=0;
    ll cnt=1;
    while(s<maxn)
    {
        f[cnt++]=s;
      //  printf("%d  ",f[cnt-1]);
        d=s;
        s=d+cnt;
    }
    ll n;
    cin>>n;
    for(ll i=1;i<cnt;i++)
    {	
        ll l=i,r=cnt,mid=(l+r)/2;
        ll A=f[l],B;
        while(l<=r)
        {
        	B=f[mid];
        	if(A+B<n)
        	{
        		l=mid+1;
        		mid=(l+r)/2;
			}
			else if(A+B>n)
			{
				r=mid-1;
				mid=(l+r)/2;
			}
			else {
				flag=1;
				printf("YES\n");
				break;
			}
		}
		if(flag==1) break;
    }
    if(flag==0) printf("NO\n");

    return 0;
}