CF1047A Little C Loves 3 I

原创于 2018-10-01 16:02:10 发布 · 280 阅读

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本文提供了一种解决CodeForces平台A题的有效算法。题目要求将一个正整数n分成三个非3的倍数的正整数之和。文章详细介绍了如何根据不同情况选择合适的数字组合，以满足题目要求。对于n模3等于0的情况，输出1,1,n-2；否则输出1,2,n-3。这是一种简洁而高效的解决方案。

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原题链接：http://codeforces.com/contest/1047/problem/A

Little C Loves 3 I

Little C loves number «3» very much. He loves all things about it.

Now he has a positive integer $n$ . He wants to split $n$ into $3$ positive integers $a, b, c$ , such that $a + b + c = n$ and none of the $3$ integers is a ultiple of $3$ . Help him to find a solution.

Input

A single line containing one integer $n (3≤n≤10^9)$ — the integer Little C has.

Output

Print $3$ positive integers $a, b, c$ in a single line, such that $a + b + c = n$ and none of them is a multiple of $3$ .

It can be proved that there is at least one solution. If there are multiple solutions, print any of them.

Examples

input

output

1 1 1

input

233

output

77 77 79

题解

如果 $n\ mod\ 3=0$ 输出 $1, 1, n - 2$ ，否则输出 $1, 2, n - 3$ 。

代码

#include<bits/stdc++.h>
using namespace std;
int n;
void in(){scanf("%d",&n);}
void ac(){n%3?printf("1 2 %d",n-3):printf("1 1 %d",n-2);}
int main(){in();ac();}