Codeforces Round #511 (Div. 2).A. Little C Loves 3 I（水题）

原创于 2018-09-22 11:20:58 发布 · 479 阅读

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本文介绍了一种算法，用于将任意正整数n拆分为三个正整数a、b、c，使得a+b+c=n且a、b、c均不为3的倍数。通过巧妙的数学逻辑，我们提供了三种不同的情况分析，确保了对于所有可能的输入都有解。算法简单高效，适用于竞赛编程和数论研究。

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A. Little C Loves 3 I

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Now he has a positive integer nn. He wants to split nn into 33 positive integers a,b,ca,b,c, such that a+b+c=na+b+c=n and none of the 33 integers is a multiple of 33. Help him to find a solution.

Input

A single line containing one integer nn (3≤n≤1093≤n≤109) — the integer Little C has.

Output

Print 33 positive integers a,b,ca,b,c in a single line, such that a+b+c=na+b+c=n and none of them is a multiple of 33.

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

Examples

input

Copy

output

Copy

1 1 1

input

Copy

output

Copy

77 77 79

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define M(a) memset(a,0,sizeof(a))
#define maxn 200010

int main()
{
    ll n;
    while(cin>>n)
    {
        if(n%3==0) cout<<1<<' '<<1<<' '<<n-2<<endl;
        else if(n%3==1) cout<<2<<' '<<1<<' '<<n-3<<endl;
        else cout<<2<<' '<<2<<' '<<n-4<<endl;
    }
    return 0;
}