Helpful Maths

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最新推荐文章于 2022-05-16 20:42:14 发布

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分类专栏： Codeforces 周赛

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本文介绍了一个简单的编程问题，旨在帮助名为Xenia的小学生能够轻松地计算只包含1、2、3这三个数字的有序数列之和。通过对输入字符串进行解析和排序，实现了数列的重新组织。

D - Helpful Maths

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 339A

Description

Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.

The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.

You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.

Input

The first line contains a non-empty string s — the sum Xenia needs to count. String s contains no spaces. It only contains digits and characters "+". Besides, string s is a correct sum of numbers 1, 2 and 3. String s is at most 100 characters long.

Output

Print the new sum that Xenia can count.

Sample Input

Input

3+2+1

Output

1+2+3

Input

1+1+3+1+3

Output

1+1+1+3+3

Input

Output

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int main()
{
    char ch[105];
    char c='+';
    int n,i,j,m;
    char a[105];
    while(gets(ch))
    {
        n=strlen(ch);
        m=n/2+1;
        for(i=0,j=0;j<n;i++,j+=2)
        a[i]=ch[j];
        sort(a,a+m);
        for(int k=0;k<m;k++)
        {
            if(k==0)
                cout<<a[k];
            else
                cout<<"+"<<a[k];
        }
        cout<<endl;
    }
    return 0;
}