Educational Codeforces Round 19 C. Minimal string

C. Minimal string

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Petya recieved a gift of a string s with length up to 105 characters for his birthday. He took two more empty strings t and u and decided to play a game. This game has two possible moves:

• Extract the first character of s and append t with this character.
• Extract the last character of t and append u with this character.

Petya wants to get strings s and t empty and string u lexigraphically minimal.

You should write a program that will help Petya win the game.

Input

First line contains non-empty string s (1 ≤ |s| ≤ 105), consisting of lowercase English letters.

Output

Print resulting string u.

Examples

input

cab

output

abc

input

acdb

output

abdc

题目大意：给一个字符串s , 和两个空串t, u;

两种操作：

1）提出s的首字母添加到t字符串的尾部

2）提出t字符串的尾部字母添加到u字符串的尾部

当s, t 为空后，所能得到的字典序最小的字符串u;

题目分析：

   记录每个字母出现的次数 

   对s字符串进行遍历，对当前遍历的位置i, 得到字母s[i] 加入到stack栈中， 根据输入的字母判断  (i+1)到结束   是否存在字母x < 栈首字母y; 

   如果没有，出栈。 反之继续进行遍历;

具体细节参见代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = (int)1e5+5;int top, ch[26];char s[maxn], Stack[maxn], u[maxn];  // u字符串存储最后的答案;bool calc(char u){  // false 表示没有找到;    int v = u-'a';    for(int i = 0; i < v; ++i){        if(ch[i] != 0) return true;    }    return false;}int main(){    scanf("%s", s);    int len = strlen(s);    for(int i = 0; i < len; ++i) ch[s[i]-'a']++;    top = 0;    int r = 0;  // r表示 字符串u当前的长度;    for(int i = 0; i < len; ++i){        ch[s[i]-'a']--;        Stack[top++] = s[i];        while(top > 0 && calc(Stack[top-1])==false){            u[r++] = Stack[--top];        }    }    u[r++] = '\0';    printf("%s", u);}