【CodeForces - 199A】【 Hexadecimal's theorem 】

版权声明：本人原创，未经许可，不得转载 https://blog.csdn.net/qq_42505741/article/details/83339841

题目：

Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.

Let's remember how Fibonacci numbers can be calculated. F0 = 0, F1 = 1, and all the next numbers are Fi = Fi - 2 + Fi - 1.

So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ...

If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number n by three not necessary different Fibonacci numbers or say that it is impossible.

Input

The input contains of a single integer n (0 ≤ n < 109) — the number that should be represented by the rules described above. It is guaranteed that n is a Fibonacci number.

Output

Output three required numbers: a, b and c. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.

If there are multiple answers, print any of them.

Examples

Input

3

Output

1 1 1

Input

13

Output

2 3 8

解题报告：水题，给定咱们一个斐波那契数，问它是否能够拆分成三个斐波那契数的和，我的做法是进行特判，当n是小于等于2

的时候只能是0 0 n 大于二的时候是可以将前一项继续拆分为两个斐波那契数的和，所以公式就是num[i]=num[i-2]+num[i-2]+num[i-3];虽然和给出的样例不是很一致，但是却符合条件，所以能A

ac代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstring>
using namespace std;

const int maxn=1e5+10000;
int num[maxn];
void db()
{
	num[0]=num[1]=1;
	for(int i=2;i<=maxn;i++)
	{
		num[i]=num[i-1]+num[i-2];
		if(num[i]>1e9)
			break;
	}
}
int main()
{
	int n;
	db();
	while(scanf("%d",&n)!=EOF)
	{
		if(n<=2)
		{
			printf("0 0 %d\n",n);
		}
		else
		{
			int i;
			for(i=1;i<n;i++)
				if(num[i]==n)
					break;
			printf("%d %d %d\n",num[i-3],num[i-2],num[i-2]);
		}
	}
}