zoukankan      html  css  js  c++  java
  • ZOJ 17届校赛 Knuth-Morris-Pratt Algorithm( 水题)

    In computer science, the Knuth-Morris-Pratt string searching algorithm (or KMP algorithm) searches for occurrences of a "word" W within a main "text string" S by employing the observation that when a mismatch occurs, the word itself embodies sufficient information to determine where the next match could begin, thus bypassing re-examination of previously matched characters.

    Edward is a fan of mathematics. He just learnt the Knuth-Morris-Pratt algorithm and decides to give the following problem a try:

    Find the total number of occurrence of the strings "cat" and "dog" in a given string s.

    As Edward is not familiar with the KMP algorithm, he turns to you for help. Can you help Edward to solve this problem?

    Input

    There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 30), indicating the number of test cases. For each test case:

    The first line contains a string s (1 ≤ |s| ≤ 1000).

    Output

    For each case, you should output one integer, indicating the total number of occurrence of "cat" and "dog" in the string.

    Sample Input

    7
    catcatcatdogggy
    docadosfascat
    dogdddcat
    catcatcatcatccat
    dogdogdogddddooog
    dcoagtcat
    doogdog
    

    Sample Output

    4
    1
    2
    5
    3
    1
    1
    

    Hint

    For the first test case, there are 3 "cat" and 1 "dog" in the string, so the answer is 4.

    For the second test case, there is only 1 "cat" and no "dog" in the string, so the answer is 1.

     水题

     1 #include<cstdio>
     2 #include<iostream>
     3 #include<algorithm>
     4 #include<cmath>
     5 #include<queue>
     6 #include<stack>
     7 #include<cstring>
     8 #include<string>
     9 using namespace std;
    10 
    11 char  a[1005];
    12 
    13 int main()
    14 {
    15     int T,len,sum,i;
    16     while(~scanf("%d",&T))
    17     {
    18         while(T--)
    19         {
    20             scanf("%s",&a);
    21             len=strlen(a);
    22             sum=0;
    23             for(i=0;i<len-2;i++)
    24             {
    25                 if(a[i]=='c'&&a[i+1]=='a'&&a[i+2]=='t')
    26                     sum++;
    27                 else if(a[i]=='d'&&a[i+1]=='o'&&a[i+2]=='g')
    28                     sum++;
    29             }
    30             printf("%d
    ",sum);
    31         }
    32     }
    33     return 0;
    34 }
  • 相关阅读:
    js设置与获取Cookie
    js,正则应用
    Ajax支持跨域之Web API实现
    RSA加密解密在jsencrypt+c#的实现-博客园加密登录
    转:sqlserver无法创建索引,超时时间已到解决办法
    【转】asp.net 项目在 IE 11 下出现 “__doPostBack”未定义 的解决办法
    c#连接SFTP上传文件
    mac 修改mysql root密码
    Vue学习手记09-mock与axios拦截的使用
    Vue学习手记08-vue-cli的启动过程
  • 原文地址:https://www.cnblogs.com/Annetree/p/6690335.html
Copyright © 2011-2022 走看看