Problem 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
代码语言：javascript复制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 1Sample Output
代码语言：javascript复制6
-1
#include <stdio.h>
int n, m;
int a[1000005];
int b[10005];
int f[10005];
void getfail()
{
f[0] = 0;
f[1] = 0;
for (int i = 1; i < m; i )
{
int j = f[i];
while (j && b[i] != b[j])
j = f[j];
f[i 1] = b[j] == b[i] ? j 1 : 0;
}
}
int main()
{
int flag, i, j, t;
scanf("%d",&t);
while (t--)
{
scanf("%d %d",&n,&m);
for (i = 0; i < n; i )
scanf("%d",&a[i]);
for (i = 0; i < m; i )
scanf("%d",&b[i]);
getfail();
for (i = 0;i <= m; i ) printf("%d ",f[i]); puts("");
flag = 1;
j = 0;
for (i = 0;i < n;i )
{
while (j && b[j] != a[i])
j = f[j];
if (b[j] == a[i])
j ;
if (j == m)
{
flag = 0;
printf("%dn",i - m 1);
}
}
if (flag)
puts("-1");
}
return 0;
}
