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
, 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
. 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\n", 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