九度OJ 1094：String Matching（字符串匹配） （计数）

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：1259

解决：686

题目描述：

    Finding all occurrences of a pattern in a text is a problem that arises frequently in text-editing programs. 
    Typically,the text is a document being edited,and the pattern searched for is a particular word supplied by the user.  
    We assume that the text is an array T[1..n] of length n and that the pattern is an array P[1..m] of length m<=n.We further assume that the elements of P and  T are all alphabets(∑={a,b...,z}).The character arrays P and T are often called strings of characters.  
    We say that pattern P occurs with shift s in the text T if 0<=s<=n and T[s+1..s+m] = P[1..m](that is if T[s+j]=P[j],for 1<=j<=m).  
    If P occurs with shift s in T,then we call s a valid shift;otherwise,we calls a invalid shift. 
    Your task is to calculate the number of vald shifts for the given text T and p attern P.

输入：

   For each case, there are two strings T and P on a line,separated by a single space.You may assume both the length of T and P will not exceed 10^6. 

输出：

    You should output a number on a separate line,which indicates the number of valid shifts for the given text T and pattern P.

样例输入：

abababab abab

样例输出：

3

来源：

2006年上海交通大学计算机研究生机试真题

思路：

简单的计数题。

代码：

#include <stdio.h>#include <string.h> #define N 1000000 int main(void){    int tlen, plen, i;    char t[N+1], p[N+1];     while (scanf("%s%s", t, p) != EOF)    {        tlen = strlen(t);        plen = strlen(p);        int count = 0;        for(i=0; i<=tlen-plen; i++)        {            if (t[i] == p[0] && strncmp(t+i, p, plen) == 0)                count ++;        }        printf("%d\n", count);    }     return 0;}/**************************************************************    Problem: 1094    User: liangrx06    Language: C    Result: Accepted    Time:30 ms    Memory:2788 kb****************************************************************/