Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8463 Accepted Submission(s): 3875
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
kmp水
#include<stdio.h> #include<string.h> int a[1000005],b[10005],next[10005]; int n,m; void get_next() { next[0]=-1; int i=0,j=-1; while(i<m-1) { if(j==-1||b[i]==b[j]) next[++i]=++j; else j=next[j]; } } int kmp() { int i=-1,j=-1; while(i<n&&j<m) { if(j==-1||a[i]==b[j]) i++,j++; else j=next[j]; } if(j==m) return i-j+1; return 0; } int main() { int t,i,ans; 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]); if(n<m) { printf("-1 "); continue; } get_next(); ans=kmp(); if(ans) { printf("%d ",ans); } else printf("-1 "); } return 0; }