A - Number Sequence
Time Limit: 5000 MS Memory Limit: 32768 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
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.
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
#include<stdio.h> using namespace std; int n,m,slen,tlen; int s[1000010]; int t[10010]; int next[10010]; void getnext() { int j,k; j=0; k=-1; next[0]=-1; while(j<tlen) { if(k==-1 || t[j]==t[k]) { next[++j] = ++k; } else { k = next[k]; } } } int kmp_index() { int i=0,j=0; getnext(); while(i<slen && j<tlen) { if(j==-1 || s[i]==t[j] ) { i++; j++; } else { j = next[j]; } } if(j == tlen) { return i-tlen+1; } else { return -1; } } int main() { int m; scanf("%d",&m); while(m--) { scanf("%d%d",&slen,&tlen); for(int i=0; i<slen; i++) { scanf("%d",&s[i]); } for(int j=0; j<tlen; j++) { scanf("%d",&t[j]); } int q = kmp_index(); printf("%d\n",q); } }