Number Sequence
Time Limit:5000MS Memory Limit:32768KB 64bit
IO Format:%I64d & %I64u
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
【一道没有不论什么杂质的纯KMP】
#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int N = 1000010; const int M = 10010; int nxt[M]; int P[N], T[M]; int n,m; void getnext(){ int j, k; j = 0; k = -1; nxt[0] = -1; while(j<m){ if (k==-1 || T[j]==T[k]){ nxt[++j] = ++k; } else{ k = nxt[k]; } } } int kmp(){ getnext(); int j, k; j = 0; k = 0; while(k<m && j<n){ if (k==-1 || T[k]==P[j]){ ++k;++j; if(k==m) return j-m+1; } else{ k = nxt[k]; } } 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",&P[i]); for(int i=0;i<m;i++) scanf("%d",&T[i]); printf("%d\n", kmp()); } return 0; }