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Codeforces Round #197 (Div. 2) A. Helpful Maths【字符串/给一个连加计算式,只包含数字 1、2、3,要求重新排序,使得连加的数字从小到大】

时间:2017-12-05 23:18:09      阅读:320      评论:0      收藏:0      [点我收藏+]

标签:lin   clu   char s   you   dex   tip   alc   nbsp   numbers   

A. Helpful Maths
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

Examples
input
3+2+1
output
1+2+3
input
1+1+3+1+3
output
1+1+1+3+3
input
2
output
2

 【代码】:

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

const int N=100005;
int k;
char s[110];
int a[110];

int main()
{
    scanf("%s",s);
    for(int i=0;i<strlen(s);i+=2)//+=2
    {
        a[k++]=s[i]-‘0‘;
    }
    sort(a,a+k);
    
    printf("%d",a[0]);//第一个前面无+,单独输出
    for(int i=1;i<k;i++)
        printf("+%d",a[i]);
    printf("\n");
}

  

Codeforces Round #197 (Div. 2) A. Helpful Maths【字符串/给一个连加计算式,只包含数字 1、2、3,要求重新排序,使得连加的数字从小到大】

标签:lin   clu   char s   you   dex   tip   alc   nbsp   numbers   

原文地址:http://www.cnblogs.com/Roni-i/p/7989963.html

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