POJ 3302 Subsequence（我的水题之路——子串判断）

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6940		Accepted: 4058

Description

Given a string s of length n, a subsequence of it, is defined as another string s' = s_u1s_u2...s_um 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

给两个字符串，str1，str2，问是否str2是str1正序或逆序的子串。

如：

1)

arash aah

a r a s h (正序)

a   a    h 是子串

2）

arash hsr

h s a r a(逆序，正序不是)

h s    r   是子串。

直接用一个循环，遍历str1，与str2进行单字符匹配，如果匹配成功就count++，如果count==strlen(str1),则输出YES，遍历正序、逆序各一次，都没有则输出NO。

代码（1AC）：

#include <cstdio>
#include <cstdlib>
#include <cstring>

char str1[110];
char str2[110];

int main(void){
    int ii, casenum;
    int count;
    int len1, len2;
    int i, j;
    int flag;

    scanf("%d", &casenum);
    getchar();
    for (ii = 0; ii < casenum; ii++){
        scanf("%s%s", str1, str2);
        len1 = strlen(str1);
        len2 = strlen(str2);
        flag = 0;
        for (i = j = count =  0; !flag && j < len1; j++){
            if (str1[j] == str2[i]){
                count++;
                i++;
            }
            if (count == len2){
                flag = 1;
            }
        }
        for (count = i = 0, j = len1 - 1; !flag && j >= 0; j--){
            if (str1[j] == str2[i]){
                count++;
                i++;
            }
            if (count == len2){
                flag = 1;
            }
        }
        if (flag){
            printf("YES\n");
        }
        else{
            printf("NO\n");
        }
    }
    return 0;
}