A. Little C Loves 3 I

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最新推荐文章于 2024-11-18 21:13:50 发布

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本文介绍了解决一道数学竞赛题的方法，题目要求将一个正整数n拆分成三个正整数a、b、c，使得它们的和等于n且任意一个数都不是3的倍数。通过分析，给出了一个有效的算法实现，适用于3到10^9之间的n。

滴答滴答---题目链接

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.这里是解题的关键

Examples

input

Copy

output

Copy

1 1 1

input

Copy

output

Copy

77 77 79

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long n;
    scanf("%lld",&n);
    long long ans1,ans2,ans3;
    long long l=n-2;
    if(l%3==0)
        ans1=1,ans2=2,ans3=l-1;
    else
        ans1=1,ans2=1,ans3=l;
    printf("%lld %lld %lld\n",ans1,ans2,ans3);
    return 0;
}