HDU 5918 Sequence I （水题+字符串匹配）Sequence I

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概述

Sequence I

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 862 Accepted Submission(s): 332

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

Source

2016中国大学生程序设计竞赛（长春）-重现赛

题意：给两个序列和一个步长，问其中一个序列是否可以在确定步长的情况下构造一个完全匹配另一个序列的子序列，如果可以统计起始位置有多少个。

思路：一开始听了zy1的思路说要用KMP，然后我就敲完模板发现自己还是不会用KMP，后来又读了一遍题，发现我都构造出来了直接判断相等就可以了..不过如果用KMP的话应该就不用提前构造了吧。

#include <iostream>
#include <string.h>
#include <algorithm> 
#include <stdio.h>
using namespace std; 
int n,m;
int a[100005];
int b;

int main(){
    
    int t;
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++){
        int q;
        scanf("%d %d %d",&n,&m,&q);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        string str="";
        for(int i=0;i<m;i++){
            scanf("%d",&b);
            str+=(b+'0');
        }
        int ans=0;
        for(int i=0;i+(m-1)*q<n;i++){
            string s="";
            for(int j=i;j<=i+(m-1)*q;j+=q){
                s+=(a[j]+'0');
            }
            if(str==s)
                ans++;
        }
        
        printf("Case #%d: %dn",cas,ans);
    }
    return 0;
}