2016 ccpc长春现场赛I Sequence I(hdu 5918) Sequence I

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我是靠谱客的博主朴实果汁，最近开发中收集的这篇文章主要介绍2016 ccpc长春现场赛I Sequence I(hdu 5918) Sequence I，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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


2
6 3 1
1 2 3 1 2 3
1 2 3
6 3 2
1 3 2 2 3 1
1 2 3

Sample Output


Case #1: 2
Case #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 100005
typedef 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: %dn", tcase, sum);
}
}