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.
InputThe 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].
OutputFor 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简单题。
代码:
#include <stdio.h> #include <stdlib.h> #include <string.h> int n,m; int a[1000001],b[10001]; void findNext(int *S,int Snum,int *Next) { int i = 0,j = -1; Next[0] = -1; while(i < Snum) { if(j == -1 || S[i] == S[j]) { Next[++ i] = ++ j; } else j = Next[j]; } } int Kmp() { int Next[10001],ans = 0; findNext(b,m,Next); int i = -1,j = -1; while(i < n) { if(j == -1 || a[i] == b[j]) { i ++,j ++; } else j = Next[j]; if(j == m) return i - m + 1; } 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]); } printf("%d ",Kmp()); } return 0; }
又做了一遍复习。
代码:
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> using namespace std; int n,m; int a[1000001],b[10001]; void getNext(int *Next) { Next[0] = -1; int i = -1,j = 0; while(j < m) { if(i == -1 || b[i] == b[j]) { Next[++ j] = ++ i; } else i = Next[i]; } } int Kmp() { int Next[10001]; getNext(Next); int i = 0,j = 0; while(j < n) { if(i == -1 || b[i] == a[j]) { i ++; j ++; if(i == m) return j - m + 1; } else i = Next[i]; } return -1; } int main() { int t; scanf("%d",&t); for(int i = 0;i < t;i ++) { scanf("%d%d",&n,&m); for(int j = 0;j < n;j ++) { scanf("%d",&a[j]); } for(int j = 0;j < m;j ++) { scanf("%d",&b[j]); } printf("%d ",Kmp()); } }