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2014嘉杰信息杯ACM/ICPC湖南程序设计邀请赛暨第六届湘潭市程序设计竞赛 E

2014-06-07 16:45 477 查看

Welcome to XTCPC
Accepted : 155		Submit : 292
Time Limit : 1000 MS		Memory Limit : 65536 KB

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

水题交错了3遍这种方法错在了遍历时没按顺序，下为AC代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;

int main()
{
    int n,i;
    int q=1;
    char a[12000];
    scanf("%d",&n);
    getchar();
    while(n--)
    {
       scanf("%s",a);

        int r1=-1,r2=-1,r3=-1,r4=-1,r5=-1;
        int k1=0,k2=0,k3=0,k4=0;
        int len=strlen(a);
        for(i=0;i<len;i++)
        {

            if(a[i]=='X'&&k1==0)
            {
                r1=i;
                k1=1;
            }
           if(a[i]=='T'&&k2==0&&k1)
            {
                r2=i;
                k2=1;
            }
            if(a[i]=='C'&&k3==0&&k1&&k2)
            {
                r3=i;
                k3=1;
            }
               if(a[i]=='P'&&k4==0&&k1&&k2&&k3)
            {
                r4=i;
                k4=1;
            }

        }

        for(i=r4+1;i<len;i++)
        {
            if(a[i]=='C')
            {
                r5=i;
                break;
            }
        }
      
        printf("Case %d: ",q++);
        if(r1!=-1&&r2!=-1&&r3!=-1&&r4!=-1&&r5!=-1)
        {
            if(r1<r2&&r2<r3&&r3<r4&&r4<r5)
               printf("Yes\n");
            else printf("No\n");
        }
        else printf("No\n");

     }

    return 0;
}

这种方法较上个方法简单

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;
char a[10000005];
int main()
{
    int n,i,j;
    char str[]={'X','T','C','P','C'};
    scanf("%d",&n);
    getchar();
    for(j=1;j<=n;j++)
    {
        int k=0;
       scanf("%s",a);
        int len=strlen(a);
        for(i=0;i<len;i++)
        {
            if(a[i]==str[k])
                k++;
        }

        printf("Case %d: ",j);
        if(k==5)
            printf("Yes\n");
        else printf("No\n");

    }

    return 0;
}

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