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 1Sample Output
6 -1
题意:找出在主串的第几位出现了模式串
题解:kmp模板题,见代码
#include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> using namespace std; const int N=1e4+10; int a[N*100],b[N],net[N]; int m,n; void getnext() { int len=m; net[0]=-1; int k=-1,j=0; while(j<len) { // if(k==-1||b[j]==b[k])//未优化 // net[++j]=++k; if(k==-1||b[j]==b[k])//优化 { j++; k++; if(b[j]!=b[k]) net[j]=k; else net[j]=net[k]; } else k=net[k]; } } int kmp() { int i=0,j=0; int lena=n,lenb=m; while(i<lena&&j<lenb) { if(j==-1||a[i]==b[j]) { i++; j++; } else j=net[j]; } if(j==lenb) return i-j+1; else return -1; } 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(); printf("%d\n",kmp()); } }