Codeforces Round #550 (Div. 3)A. Diverse Strings

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A. Diverse Strings time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output A string is called diverse if it contains consecutive (adjacent) letters of the Latin alphabet and each letter occurs exactly once. For example, the following strings are diverse: "fced", "xyz", "r" and "dabcef". The following string are not diverse: "az", "aa", "bad" and "babc". Note that the letters 'a' and 'z' are not adjacent.

Formally, consider positions of all letters in the string in the alphabet. These positions should form contiguous segment, i.e. they should come one by one without any gaps.

You are given a sequence of strings. For each string, if it is diverse, print “Yes”. Otherwise, print “No”.

Input
The first line contains integer n (1≤n≤100), denoting the number of strings to process. The following n lines contains strings, one string per line. Each string contains only lowercase Latin letters, its length is between 1 and 100, inclusive.

Output
Print n lines, one line per a string in the input. The line should contain “Yes” if the corresponding string is diverse and “No” if the corresponding string is not diverse. You can print each letter in any case (upper or lower). For example, “YeS”, “no” and “yES” are all acceptable.

Example
input

8
fced
xyz
r
dabcef
az
aa
bad
babc

output

Yes
Yes
Yes
Yes
No
No
No
No

题意：给定字符串，判断字符串中的字母是否只出现一次并且出现的字母是连续的。
思路：排序，证明连续；
参考代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define N 1005
using namespace std;
int ans[N];
char sum[40];
int main()
{
    int n;
    string a;
    scanf("%d",&n);
    while(n--)
    {
        memset(ans,0,sizeof(ans));
        cin>>a;
        int maxx = 0,minn ='z'-'a';
        for(int i=0; i<a.length(); ++i)
        {
            ans[a[i]-'a']++;
            maxx = max(maxx,a[i]-'a');
            minn = min(minn,a[i]-'a');
        }
        bool flag =0;
        for(int i=minn; i<=maxx; ++i)
        {
            if(ans[i] != 1)
            {
                flag =1;
                break;
            }
        }
        if(flag)
            printf("No\n");
        else
            printf("Yes\n");
    }
    return 0;
}