CodeForces 672B Different is Good

你的名字 2022-08-22 12:21 32阅读 0赞

# CodeForces 672B Different is Good #

time limit per test:2 seconds  
memory limit per test:256 megabytes  
input:standard input  
output:standard output

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A wise man told Kerem “Different is good” once, so Kerem wants all things in his life to be different.

Kerem recently got a string s consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string s to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string “aba” has substrings “” (empty substring), “a”, “b”, “a”, “ab”, “ba”, “aba”.

If string s has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.

Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.

## Input ##

The first line of the input contains an integer n (1 ≤ n ≤ 100 000) — the length of the string s.

The second line contains the string s of length n consisting of only lowercase English letters.

## Output ##

If it’s impossible to change the string s such that all its substring are distinct print -1. Otherwise print the minimum required number of changes.

## Examples ##

## Input ##

> 2  
> aa

## Output ##

> 1

## Input ##

> 4  
> koko

## Output ##

> 2

## Input ##

> 5  
> murat

## Output ##

> 0

## Note ##

In the first sample one of the possible solutions is to change the first character to ‘b’.

In the second sample, one may change the first character to ‘a’ and second character to ‘b’, so the string becomes “abko”.

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题目在这 [题目链接][Link 1]

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如果有相同的字母，把他们换成未出现过的不同字母，最少换几位  
map记录下出现次数就行了

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代码：

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #include<map>
    using namespace std;
    
    int main()
    {
        int n;
        while(~scanf("%d",&n))
        {
            char a[100005];
            scanf("%s",a);
            if(n>26)
            {
                printf("-1\n");
            }
            else
            {
                int ans=0;
                map<char,int> ch;
                for(int i=0;i<n;i++)
                {
                    if(ch[a[i]]==0)
                    {
                        ch[a[i]]=1;
                    }
                    else
                    {
                        ans++;
                    }
                }
                printf("%d\n",ans);
            }
    
        }
        return 0;
    }

[Link 1]: http://codeforces.com/problemset/problem/672/B