POJ 1963：All in All

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Description

Input

Output

Sample Input

sequence subsequence
person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter

Sample Output

Yes
No
Yes
No

简单的字符串问题。。判断后一个字符串是否包含前一个。。好久没0MS  1A 了 。。。OTL。。。

#include<iostream>
#include<cstring>
#include<ctime>
#include<algorithm>
#include<cstdio>
#include<queue>
using namespace std;

const int maxn = 100000 + 50;
char a[maxn];
char b[maxn];

int main()
{
    while(scanf("%s%s", a, b)==2)
    {
        int x = strlen(a);
        int y = strlen(b);
        int t = 0;
        int judge = 0;
        if( y<x )
        {
            printf("No\n");
            continue;
        }
        for(int j=0; j<y; j++)
        {
            if( b[j] == a[t] )
            {
                t++;
                judge++;
            }
        }
        if( judge==x )
            printf("Yes\n");
        else
            printf("No\n");
    }

    return 0;
}

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POJ 1963：All in All

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原文地址：http://blog.csdn.net/u013487051/article/details/38178913