ZZULI_SummerPractice(4) POJ 3302…

最新推荐文章于 2020-08-04 16:55:33 发布

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本文介绍了一个简单的算法，用于检查一个字符串是否为另一个字符串的子序列，或是其反转字符串的子序列。通过两个输入字符串，该算法能有效地判断并输出结果。

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Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6570		Accepted: 3829

Description

Given a string s of length n, a subsequence of it, is defined as another string s' = s_u_₁s_u_₂...s_{u_m} where 1 ≤ u₁ < u₂ < ... < u_m ≤ n and s_i is the ith character of s. Your task is to write a program that, given two strings s1 and s2, checks whether either s2 or its reverse is a subsequence of s1 or not.

Input

The first line of input contains an integer T, which is the number of test cases. Each of the next T lines contains two non-empty strings s1 and s2 (with length at most 100) consisted of only alpha-numeric characters and separated from each other by a single space.

Output

For each test case, your program must output "YES", in a single line, if either s2 or its reverse is a subsequence of s1. Otherwise your program should write "NO".

Sample Input

5
arash aah
arash hsr
kick kkc
A a
a12340b b31

Sample Output

YES
YES
NO
NO
YES

Source

Amirkabir University of Technology Local Contest 2006

题比较简单，就是从第一个字符串中按顺序看看能不能找到第二个字符串的字母，或者第二个字符串倒置后的字母。

代码：

#include<stdio.h>

#include<string.h>

int main()

{

int n,i,j,len;

char str1[102],str2[102],str[102];

scanf("%d",&n);

while(n--)

{

scanf("%s%s",str1,str2);

len=strlen(str1);

j=0;

for(i=len-1;i>=0;i--)

str[j++]=str1[i];

str[j]=0;

i=j=0;

while(str2[i]!=0)

{

if(str1[j]==str2[i]){i++;j++;}

else j++;

if(str1[j]==0)break;

}

if(str2[i]==0)

printf("YES\n");

else{

i=j=0;

while(str2[i]!=0)

{

if(str[j]==str2[i]){i++;j++;}

else j++;

if(str[j]==0)break;

}

if(str2[i]==0)printf("YES\n");

else printf("NO\n");

}

return 0;

}

ZZULI_SummerPractice(4)&nbsp;POJ&nbsp;3302…

ZZULI_SummerPractice(4) POJ 3302…