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[], ch2[];

 int main(){

     int n, a, b;

     while(~scanf("%s%s", ch1, ch2)){

         int l1 = strlen(ch1);

         int l2 = strlen(ch2);

         a = ;

         b = l2-;

         for(int i = ; i < l1 && i < l2; i++){

             if(ch1[i] == ch2[i]){

                 a = i+;

                 ch1[i] = '#'; //避免重复判断

             }

             else

                 break;

         }

         for(int i = ; i <= l1&&i <= l2;i++){

             if(ch1[l1-i] == ch2[l2-i]){

                 b = l2-i-;

             }

             else

                 break;

         }

         if(a > b)

             printf("0\n");

         else

             printf("%d\n", b-a+);

     }

     return ;

 }