csuoj 1394: Virus Replication

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1394

1394: Virus Replication

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 135  Solved: 40
[Submit][Status][Web Board]

Description

Input

Output

Sample Input

AAAAA
AGCGAA

Sample Output

3

HINT

Source

分析；
题目意思是找出在第一个串中第二个串没有出现的字母个数，要求从前往后遍历和从后往前遍历两次。

AC代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#include<stack>
#include<map>
#include<string>
using namespace std;
char ch1[100050], ch2[100050];
int main(){
    int n, a, b;
    while(~scanf("%s%s", ch1, ch2)){
        int l1 = strlen(ch1);
        int l2 = strlen(ch2);
        a = 0;
        b = l2-1;
        for(int i = 0; i < l1 && i < l2; i++){
            if(ch1[i] == ch2[i]){
                a = i+1;
                ch1[i] = '#'; //避免重复判断 
            }
            else
                break;
        }
        for(int i = 1; i <= l1&&i <= l2;i++){
            if(ch1[l1-i] == ch2[l2-i]){
                b = l2-i-1;
            }
            else
                break; 
        }
        if(a > b)
            printf("0
");
        else
            printf("%d
", b-a+1);
    }
    return 0;
}