CodeForces - 192A

最新推荐文章于 2025-07-03 16:13:32 发布

原创最新推荐文章于 2025-07-03 16:13:32 发布 · 389 阅读

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#算法 #数据结构

本文介绍了一种算法，用于判断给定的整数是否可以表示为两个三角数之和。核心在于利用数学技巧和编程实现，通过迭代寻找符合条件的三角数组合。挑战者需帮助Andrew解决趣味与酷炫的数学问题。

A. Funky Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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

int main()
{
	int i=1,n,k,f=0,m1;
	long long int sum1,sum2,sum,x,sum3=0,j=1;
	scanf("%d",&n);

	sum2=j*(j+1)/2;
	while(1)
	{
		j=1;
		sum1=i*(i+1)/2;
		sum=n-sum1*2;
		//printf("%d\n",sum);//
		x=i+1;
		//printf("%d\n",x);//
		if(sum==0)
		{
			f=1;
			break;
			
		}
		else if(sum<0){
			break;
		}
		else
		{
			while(sum-sum3>0)
			{				
				sum3=(x+x+j-1)*j/2;
				if(sum-(x+x+(j+10000)-1)*(j+10000)/2>10)//优化速度，要不然会超时，【哭死】
				{
					j=j+10000;
				}
				else
				j++;
				//printf("%d\n",sum3);//
			}
				//printf("%d\n",sum-sum3);//
			if(sum-sum3==0)
			{
				f=1;
				break;
			}
		}
		i++;sum3=0;
	}
	if(f==1)
	printf("YES");
	else
	printf("NO");
	//printf("\n%d",i);
	//printf("\n");
	/*for(j=10;j<100;j++)
	{
		printf("%d\n",j*(j+1)/2);
	}*/
	return 0;
}