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 ;
 }