#C++初学记录(ACM试题1）

**A - Diverse Strings **

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. And all letters in the string should be distinct (duplicates are not allowed).

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

正确代码

#include<cstdio>
#include<cstring>
int main(){
    char a[105];
    int t;
    scanf("%d", &t);
    while(t--){
        memset(a, 0, sizeof(a));
        scanf("%s", a);
        int len = strlen(a), flag = 0;
        for(int i = 0; i < len; i++){
            for(int j = i+1; j < len; j++){
                if(a[i] > a[j]){
                    char temp = a[i];
                    a[i] = a[j];
                    a[j] = temp;
                }
            }
        }
        for(int i = 1; i < len; i++){
            if(a[i] - a[i-1] != 1){
                printf("No\n");
                flag = 1;
                break;
            }
        }
        if(flag == 1){
            continue;
        }else{
            printf("Yes\n");
        }
    }
    return 0;
}

题目理解

该题较为简单，该题的正确做法是对输入的字母进行判断，若有相近的字母则输出YES，若没有则输出NO，若输入的是单个字符则也输出YES。

相关知识点

%s是进行字符串的整体输入，即

int c[200];
scanf("%s",%c);

是将整个字符串拆分成单个字符储存进数组a，即输入abcde则a[1]=b。以及对单个字符进行判断，判断只需要在判断NO和YES是加一点小聪明.

        for(int i = 1; i < len; i++){
            if(a[i] - a[i-1] != 1){
                printf("No\n");
                flag = 1;
                break;
            }
        }
        if(flag == 1){
            continue;
        }else{
            printf("Yes\n");

在这个if条件判断中，因为单个字符的长度len=1，所以在进行No的判断中是不符合的，因为No的判断条件是i=1;i<len；因此直接进行else循环。