Codeforces 196D Lexicographically Maximum Subsequence（模拟+哈希）

￡神魔★判官ぃ 2022-06-02 05:25 152阅读 0赞

You've got string *s*, consisting of only lowercase English letters. Find its lexicographically maximum subsequence.

We'll call a non-empty string *s*\[*p*1*p*2... *p**k*\] = *s**p*1*s**p*2... *s**p**k*(1 ≤ *p*1 < *p*2 < ... < *p**k* ≤ |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|.

String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*|, if either |*x*| > |*y*| and *x*1 = *y*1, *x*2 = *y*2, ... , *x*|*y*| = *y*|*y*|, or exists such number *r* (*r* < |*x*|, *r* < |*y*|), that *x*1 = *y*1, *x*2 = *y*2, ... , *x**r* = *y**r* and *x**r* + 1 > *y**r* + 1. Characters in lines are compared like their ASCII codes.

Input

The single line contains a non-empty string *s*, consisting only of lowercase English letters. The string's length doesn't exceed 105.

Output

Print the lexicographically maximum subsequence of string *s*.

Example

Input

ababba

Output

bbba

Input

abbcbccacbbcbaaba

Output

cccccbba

Note

Let's look at samples and see what the sought subsequences look like (they are marked with uppercase bold letters).

The first sample: aBaBBA

The second sample: abbCbCCaCbbCBaaBA

题解：

很简单的一题，直接从最大的那个字符开始输出，用哈希存一下这之后所有字符的情况，每次直到最大的才输出就好了

代码：

#include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<queue>
    #include<stack>
    #include<math.h>
    #include<vector>
    #include<map>
    #include<set>
    #include<stdlib.h>
    #include<cmath>
    #include<string>
    #include<algorithm>
    #include<iostream>
    #include<stdio.h>
    using namespace std;
    #define ll long long
    int a[30];
    int b[30];
    int main()
    {
        char s[100005];
        int i,j,maxx=0,d=0,len;
        scanf("%s",s);
        memset(a,0,sizeof(a));
        len=strlen(s);
        for(i=0;s[i];i++)
        {
            if(maxx<s[i]-'a')
            {
                d=i;
                maxx=s[i]-'a';
                memset(a,0,sizeof(a));
            }
            a[s[i]-'a']++;
        }
        for(i=d;i<len;i++)
        {
            if(maxx==(s[i]-'a'))
                printf("%c",s[i]);
            a[s[i]-'a']--;
            if(a[maxx]==0)
            {
                while(maxx>0)
                {
                    maxx--;
                    if(a[maxx])
                    {
                        break;
                    }
                }
            }
        }
        printf("\n");
        return 0;
    }