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

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：

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<queue>
 5 #include<iostream>
 6 #include<stack>
 7 #include<map>
 8 #include<string>
 9 using namespace std;
10 char ch1[100050], ch2[100050];
11 int main(){
12     int n, a, b;
13     while(~scanf("%s%s", ch1, ch2)){
14         int l1 = strlen(ch1);
15         int l2 = strlen(ch2);
16         a = 0;
17         b = l2-1;
18         for(int i = 0; i < l1 && i < l2; i++){
19             if(ch1[i] == ch2[i]){
20                 a = i+1;
21                 ch1[i] = '#'; //避免重复判断 
22             }
23             else
24                 break;
25         }
26         for(int i = 1; i <= l1&&i <= l2;i++){
27             if(ch1[l1-i] == ch2[l2-i]){
28                 b = l2-i-1;
29             }
30             else
31                 break; 
32         }
33         if(a > b)
34             printf("0
");
35         else
36             printf("%d
", b-a+1);
37     }
38     return 0;
39 }