CodeForces - 946C String Transformation(读懂题意+暴力)

C. String Transformation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string s consisting of |s| small english letters.

In one move you can replace any character of this string to the next character in alphabetical order (a will be replaced with b, s will be replaced with t, etc.). You cannot replace letter z with any other letter.

Your target is to make some number of moves (not necessary minimal) to get string abcdefghijklmnopqrstuvwxyz (english alphabet) as a subsequence. Subsequence of the string is the string that is obtained by deleting characters at some positions. You need to print the string that will be obtained from the given string and will be contain english alphabet as a subsequence or say that it is impossible.

Input

The only one line of the input consisting of the string s consisting of |s| (1 ≤ |s| ≤ 105) small english letters.

Output

If you can get a string that can be obtained from the given string and will contain english alphabet as a subsequence, print it. Otherwise print «-1» (without quotes).

Examples

input

Copy

aacceeggiikkmmooqqssuuwwyy

output

Copy

abcdefghijklmnopqrstuvwxyz

input

Copy

thereisnoanswer

output

Copy

-1

题意：给你一个字符串，长度n<=1e5，你要把其中的的某些字母换成大于等于它的任意一个字母（例如a可以换成b到z，不换也可以），使最终这个序列中从开始到最后依次包含abcdefghijklmnopqrstuvwxyz。最后输出的是经修改过后的包含a到z的原字符串。

思路：感觉这道题就是看你能不能读懂题。。。

直接暴力求即可。。。

代码：

#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=100010;
int n,m,k,f,flag;
int ans,tmp,cnt,dp[maxn];
int a[maxn],c[maxn];
char s[maxn];
char ch;
int main()
{
    int T,cas=1;
    while(scanf("%s",s)!=EOF)
    {
        ans=0;    flag=0;
        ch='a';
        f=0;
        int l=strlen(s);
        for(k=0;k<l;k++) {
            if(s[k]<=ch){
                s[k]=ch;
                ch++;ans++;
            }
            if(ans==26) {flag=1;break;}
        }
        if(flag) printf("%s\n",s);
        else puts("-1");
    }
    return 0;
}