zoukankan      html  css  js  c++  java
  • Good or Bad LightOJ

    A string is called bad if it has 3 vowels in a row, or 5 consonants in a row, or both. A string is called good if it is not bad. You are given a string s, consisting of uppercase letters ('A'-'Z') and question marks ('?'). Return "BAD" if the string is definitely bad (that means you cannot substitute letters for question marks so that the string becomes good), "GOOD" if the string is definitely good, and "MIXED" if it can be either bad or good.

    The letters 'A', 'E', 'I', 'O', 'U' are vowels, and all others are consonants.

    Input

    Input starts with an integer T (≤ 200), denoting the number of test cases.

    Each case begins with a non-empty string with length no more than 50.

    Output

    For each case of input you have to print the case number and the result according to the description.

    Sample Input

    5

    FFFF?EE

    HELLOWORLD

    ABCDEFGHIJKLMNOPQRSTUVWXYZ

    HELLO?ORLD

    AAA

    Sample Output

    Case 1: BAD

    Case 2: GOOD

    Case 3: BAD

    Case 4: MIXED

    Case 5: BAD

    详解:http://www.cnblogs.com/jianglangcaijin/archive/2012/10/12/2721156.html

     1 #include<cstdio>
     2 #include<cstring>
     3 #include<iostream>
     4 #include<algorithm>
     5 using namespace std;
     6 
     7 int n;
     8 int a[60],d1[60][6],d2[60][6];
     9 char S[60];
    10 
    11 int OK(char x){
    12     return x=='A'||x=='E'||x=='I'||x=='O'||x=='U';
    13 }
    14 
    15 void Inite(){
    16     memset(d1,0,sizeof(d1));
    17     memset(d2,0,sizeof(d2));
    18     d1[0][0]=d2[0][0]=1;
    19     for(int i=1;i<=n;i++){
    20         if(S[i]=='?') a[i]=2;
    21         else if(OK(S[i])) a[i]=0;
    22         else a[i]=1;
    23     }
    24 }
    25 
    26 void solve(){
    27     for(int i=1;i<=n;i++){
    28         for(int j=0;j<=2;j++) if(d1[i-1][j])
    29         {    
    30             if(a[i]==2||a[i]==1) d2[i][1]=1;
    31             if(a[i]==2||a[i]==0) d1[i][j+1]=1;
    32         }
    33         for(int j=0;j<=4;j++) if(d2[i-1][j])
    34         {
    35             if(a[i]==2||a[i]==0) d1[i][1]=1;
    36             if(a[i]==2||a[i]==1) d2[i][j+1]=1;
    37         }
    38     }
    39     
    40     int bad=0,good=0;
    41     for(int i=0;i<=2;i++) if(d1[n][i]) good=1;
    42     for(int i=0;i<=4;i++) if(d2[n][i]) good=1;
    43     for(int i=1;i<=n;i++) if(d1[i][3]||d2[i][5]) bad=1;
    44     if(good&&bad) puts("MIXED");
    45     else if(good) puts("GOOD");
    46     else puts("BAD");
    47 }
    48 
    49 int main()
    50 {   int kase;
    51     cin>>kase;
    52     for(int t=1;t<=kase;t++){
    53         scanf("%s",S+1);
    54         n=strlen(S+1);
    55         printf("Case %d: ",t);
    56         Inite();
    57         solve();
    58     }
    59     return 0;
    60 } 
  • 相关阅读:
    软件定义网络实验4:Open vSwitch 实验——Mininet 中使用 OVS 命令(实验过程及结果记录)
    软件定义网络实验3:测量路径的损耗率 (实验过程及结果记录)
    第一次个人编程作业
    软件定义网络实验2:Mininet拓扑的命令脚本生成(实验过程及结果记录)
    软件定义网络实验1:Mininet源码安装和可视化拓扑工具(实验过程及结果记录)
    第一次博客作业
    第07组(69) 需求分析报告
    第七组(69)团队展示
    第三次作业
    结对编程作业
  • 原文地址:https://www.cnblogs.com/zgglj-com/p/7512794.html
Copyright © 2011-2022 走看看