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
****************************************************************/