标签：ext 简单 and clu describe printf des esc ane
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[1000000],b[10000]; 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[10000],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\n",Kmp()); } return 0; }
标签：ext 简单 and clu describe printf des esc ane
原文地址：https://www.cnblogs.com/8023spz/p/9650962.html
