http://acm.hust.edu.cn/vjudge/contest/view.action?cid=92486#problem/A

Number Sequence

Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit 

Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one. 

 

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000]. 

 

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead. 

 

Sample Input

2

 

 

13 5

1 2 1 2 3 1 2 3 1 3 2 1 2

1 2 3 1 3

 

 

13 5

1 2 1 2 3 1 2 3 1 3 2 1 2

1 2 3 2 1

 

Sample Output

6

-1

题意：在a串中找b串的位置，多个输出最小or输出-1

最简单kmp……wa哭N次，最后在大牛下终于1a

1 #include
          
            2 #include
           
             3 #include
            
              4  5 using namespace std; 6  7 #define maxn 10008 8  9 int n, m, a[maxn*100], b[maxn], Next[maxn];10 11 void getnext()12 {13     int j, k;14     j = 0;15     k = Next[0] = -1;16     while(j < m)17     {18         if(k == -1 || b[j] == b[k])19         {20             j++;21             k++;22             Next[j] = k;23         }24         else25             k = Next[k];26     }27 }28 29 void slove()30 {31     int j;32     j = 0;33     int ans = -1;34     for(int i = 0; i < n; )   //  i  不该加就不要加，你用while多好……=。=||35     {36         while(j == -1 || (a[i] == b[j] && i < n && j < m))37             i++, j++;38         if(j == m)39         {40             ans = i - j + 1;41             break;42         }43         j = Next[j];44     }45     printf("%d\n", ans);46 }47 48 int main()49 {50     int t;51     scanf("%d", &t);52     while(t--)53     {54         scanf("%d%d", &n, &m);55         for(int i = 0; i < n; i++)56             scanf("%d", &a[i]);57         for(int i = 0; i < m; i++)58             scanf("%d", &b[i]);59         getnext();60         slove();61     }62     return 0;63 }
            
           
          

刘大大说要用函数……无论多短的代码    

说的是