CF 1144A 思维

最新推荐文章于 2021-07-09 16:14:53 发布

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最新推荐文章于 2021-07-09 16:14:53 发布

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本文介绍了一种用于判断字符串是否为多样性的算法，并提供了完整的C++实现代码。所谓多样性字符串需包含连续的拉丁字母且每个字母恰好出现一次。文章通过实例详细解释了多样性的概念，并展示了如何通过排序和比较相邻字符来检查输入字符串是否符合这一特性。

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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.
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
代码：

#include<bits/stdc++.h>
using namespace std;
inline int read(){
	int x=0,f=0;char ch=getchar();
	while(ch<'0'||ch>'9')f|=ch=='-',ch=getchar();
	while(ch>='0'&&ch<='9')x=x*10+(ch^48),ch=getchar();
	return f?-x:x;
}
char s[105];
int main(){
	int t=read();
	while(t--){
		gets(s);
		int l=strlen(s),ok=1;
		sort(s,s+l);
		for(int i=1;i<l;++i){
			if(s[i]==s[i-1]+1)continue;
			puts("No");
			ok=0;
			break;
		}
		if(ok)puts("Yes");
	}
    	return 0;
}