Number Sequence

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 Status

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

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

#define maxn 10008

int n, m, a[maxn*100], b[maxn], Next[maxn];

void getnext()
{
    int j, k;
    j = 0;
    k = Next[0] = -1;
    while(j < m)
    {
        if(k == -1 || b[j] == b[k])
        {
            j++;
            k++;
            Next[j] = k;
        }
        else
            k = Next[k];
    }
}

void slove()
{
    int j;
    j = 0;
    int ans = -1;
    for(int i = 0; i < n; )   //  i  不该加就不要加，你用while多好……=。=||
    {
        while(j == -1 || (a[i] == b[j] && i < n && j < m))
            i++, j++;
        if(j == m)
        {
            ans = i - j + 1;
            break;
        }
        j = Next[j];
    }
    printf("%d
", ans);
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        for(int i = 0; i < n; i++)
            scanf("%d", &a[i]);
        for(int i = 0; i < m; i++)
            scanf("%d", &b[i]);
        getnext();
        slove();
    }
    return 0;
}

刘大大说要用函数……无论多短的代码    

说的是