http://acm.hdu.edu.cn/showproblem.php?pid=1711

##Problem Description

Given two sequences of numbers : a, a, …… , a[N], and b, b, …… , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b, a[K + 1] = b, …… , 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, a, …… , a[N]. The third line contains M integers which indicate b, b, …… , 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

### KMP的使用

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

//AC

//KMP
using namespace std;

int q;
int sub;
int nextsub;
int t, n, m;

int main(int argc, char const *argv[])
{
scanf("%d", &t);

while (t--) {

scanf("%d%d", &n, &m);

for (int i = 0; i < n; i++) {
scanf("%d", q + i);
}
for (int i = 0; i < m; i++) {
scanf("%d", sub + i);
}

nextsub = 0;

for (int i = 1; i < m; i++) {
int cpi = i;
while (1) {
if (cpi == 0) {
nextsub[i + 1] = 0;
break;
}

if (sub[nextsub[cpi]] == sub[cpi]) {
nextsub[i + 1] = nextsub[cpi] + 1;
break;
} else {
cpi = nextsub[cpi];
}
}
}

int i = 0, j = 0, find = 0;
while (i < n) {
if (q[i] == sub[j]) {
i++;
j++;

if (j == m) {
find = 1;
printf("%d\n", i - m + 1);
break;
}
} else {
if (j == 0) {
i++;
}

j = nextsub[j];
}
}
if (!find) {
printf("-1\n");
}
}

return 0;
}
``````