XTU OJ 1207 Welcome to XTCPC (字符串签到题)

本文介绍了一道简单的字符串匹配问题,旨在通过删除字符来判断一个字符串是否能转化为特定目标字符串XTCPC。该问题通过一个简洁的C语言程序解决,并提供了一个样例输入输出以验证算法的有效性。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Problem Description

Welcome to XTCPC! XTCPC start today, you are going to choose a slogan to celebrate it, many people give you some candidate string about the slogan, but the slogan itself must have something relavant to XTCPC, a string is considered relevant to XTCPC if it become XTCPC after deleting some characters in it. For example, XTCPC, XTCCPCC, OIUXKKJATSADCASPHHC is relevant, XX,FF,GG,CPCXT,XTCP is not. Now you have to write a program to judge whether a string is relevant to XTCPC.

Input

First line an integer t(t≤100), the number of testcases. For each case, there is a string(length≤100, all are uppercase characters).

Output

For each case, output case number first, then "Yes" if the string is relevant, "No" if the string is not relevant. Quote for clarify.

Sample Input
3
XTCPC
CCC
XXXXTTTTCCCCPPPCCC


Sample Output
Case 1: Yes
Case 2: No
Case 3: Yes

签到题,一道字符串的水题。这样的题目还是要争取1A,不要耗费太多的时间;
#include <cstdio>
#include <cstring>
char s[101];
char c[6]={'X','T','C','P','C'};
int main()
{
    int t,i,j,id;
    scanf("%d",&t);
    for(i=1;i<=t;i++)
    {
        id=0;
        scanf("%s",s);
        for(j=0;j<strlen(s);j++)
        {
            if(s[j]==c[id])
                id++;
        }
        if(id==5)
            printf("Case %d: Yes\n",i);
        else
            printf("Case %d: No\n",i);
    }
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值