2016 ccpc长春现场赛I Sequence I(hdu 5918)

Sequence I

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

Mr. Frog has two sequences a1,a2,⋯,an and b1,b2,⋯,bm and a number p. He wants to know the number of positions q such that sequence b1,b2,⋯,bm is exactly the sequence aq,aq+p,aq+2p,⋯,aq+(m−1)p where q+(m−1)p≤n and q≥1.

 

Input

The first line contains only one integer T≤100, which indicates the number of test cases.

Each test case contains three lines.

The first line contains three space-separated integers 1≤n≤106,1≤m≤106 and 1≤p≤106.

The second line contains n integers a1,a2,⋯,an(1≤ai≤109).

the third line contains m integers b1,b2,⋯,bm(1≤bi≤109).

 

Output

For each test case, output one line “Case #x: y”, where x is the case number (starting from 1) and y is the number of valid q’s.

 

Sample Input

26 3 11 2 3 1 2 31 2 36 3 21 3 2 2 3 11 2 3

 

Sample Output

Case #1: 2Case #2: 1

 

Statistic | Submit | Clarifications | Back题意：给你两个数组，再给你距离间隔，要求你用第一个数组以一定的距离间隔匹配第二个数组，问有多少种匹配方案。

思路：现场赛的时候这道题可以暴力O（n*m）过的，但估计这道题想考的是kmp，把kmp模板改一下就好了，kmp的时间复杂度是O（n），别问我为什么暴力比kmp速度更快。下面给代码

#include<set>#include<map>#include<ctime>#include<cmath>#include<stack>#include<queue>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<functional>typedef long long LL;using namespace std;#define inf 0x3f3f3f3f#define maxn 100005typedef long long LL;int num, sum, n, m;int next1[maxn], P[maxn], T[maxn];void makeNext(){int q, k;next1[0] = 0;for (q = 1, k = 0; q < m; ++q){while (k > 0 && P[q] != P[k])k = next1[k - 1];if (P[q] == P[k]){k++;}next1[q] = k;}}void kmp(){int i, q;makeNext();for (int st = 0; st < num; st++){for (i = st, q = 0; i < n; i += num){while (q > 0 && P[q] != T[i])q = next1[q - 1];if (P[q] == T[i]){q++;}if (q == m){sum++;}}}}int main(){int t;scanf("%d", &t);for (int tcase = 1; tcase <= t; tcase++){memset(P, 0, sizeof(P));memset(T, 0, sizeof(T));scanf("%d%d%d", &n, &m, &num);for (int i = 0; i<n; i++){scanf("%d", &T[i]);}for (int i = 0; i<m; i++){scanf("%d", &P[i]);}sum = 0;kmp();printf("Case #%d: %d\n", tcase, sum);}}