Number Sequence
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
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
题目大意：给定两个序列a，b，求出b序列在a序列中第一次出现的位置，若不在a序列中输出-1；
思路：KMP模板直接套用，这里觉得这位博主写的挺通俗易懂的还是很简单的就看懂了https://blog.****.net/starstar1992/article/details/54913261/
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string> using namespace std;
const int maxn = ;
int rs[][], nex[maxn], a[maxn], b[maxn/ + ], n, m;
void cal_next(int *str, int len)
{
nex[] = -; int k = -;
for (int i = ; i <= len - ; i++) {
while (k > - && str[k + ] != str[i])
k = nex[k];
if (str[k + ] == str[i])k = k + ;
nex[i] = k;
}
}
int KMP(int *a, int la, int *b, int lb)
{
cal_next(b , lb);
int k = -;
for (int i = ; i < la; i++) {
while (k > - && b[k + ] != a[i])
k = nex[k];
if (b[k + ] == a[i])k = k + ;
if (k == lb - )return i - lb + ;
}
return -;
}
int main()
{
ios::sync_with_stdio(false);
int T;
cin >> T;
while (T--) {
cin >> n >> m;
for (int i = ; i < n; i++)cin >> a[i];
for (int i = ; i < m; i++)cin >> b[i];
int ans = KMP(a, n, b, m);
if (ans != -)cout << ans + << endl;
else cout << "-1" << endl;
}
return ;
}