hdu 1022 AND toj 1036 栈的运用

其实就是模拟一下栈啦。

hdu 1022：

 1 #include <iostream>
 2 using namespace std;
 3 
 4 const int N = 10;
 5 char o1[N];
 6 char o2[N];
 7 char s[N];
 8 int ans[N * 2];
 9 
10 int main ()
11 {
12     int n;
13     while ( cin >> n )
14     {
15         cin >> o1 >> o2;
16         int top = 0, p = 0, cnt = 0;
17         for ( int i = 0; i < n; i++ )
18         {
19             if ( top == 0 )
20             {
21                 s[top++] = o1[p++];
22                 ans[cnt++] = 1;
23             }
24             while ( s[top - 1] != o2[i] )
25             {
26                 if ( p == n ) break;
27                 s[top++] = o1[p++];
28                 ans[cnt++] = 1;
29             }
30             if ( s[top - 1] != o2[i] )
31             {
32                 break;
33             }
34             else
35             {
36                 top--;
37                 ans[cnt++] = 0;                
38             }
39         }
40         if ( cnt == 2 * n )
41         {
42             cout << "Yes." << endl;
43             for ( int i = 0; i < cnt; i++ )
44             {
45                 cout << ( ans[i] ? "in" : "out" ) << endl;
46             }
47         }
48         else
49         {
50             cout << "No." << endl;
51         }
52         cout << "FINISH" << endl;
53     }
54     return 0;
55 }

 toj 1036（和上一个题基本一样，输入输出麻烦点）：

 1 #include <cstdio>
 2 
 3 const int N = 1000;
 4 int s[N];
 5 
 6 int main ()
 7 {
 8     int n;
 9     while ( scanf("%d", &n), n )
10     {
11         int tmp;
12         while ( scanf("%d", &tmp), tmp )
13         {
14             int top = 0, p = 1;
15             bool flag = true;
16             for ( int i = 0; i < n; i++ )
17             {
18                 if ( top == 0 )
19                 {
20                     s[top++] = p++;
21                 }
22                 while ( s[top - 1] != tmp )
23                 {
24                     if ( p == n + 1 ) break;
25                     s[top++] = p++;
26                 }
27                 if ( s[top - 1] != tmp )
28                 {
29                     flag = false;
30                 }
31                 else
32                 {
33                     top--;
34                 }
35                 if ( i != n - 1 ) 
36                 {
37                     scanf("%d", &tmp);
38                 }
39             }
40             printf("%s\n", flag ? "Yes" : "No");
41         }
42         putchar('\n');
43     }
44     return 0;
45 }

转载于:https://www.cnblogs.com/huoxiayu/p/4420834.html