hdu 5918(强行水过去..正解KMP)

Sequence I

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 216    Accepted Submission(s): 93

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:给定数组 a1...an ,b1...bm, 以及间隔 d,问 a 中使得存在 aq ,aq+d , aq+2d...aq+(m-1)d  和 b数组相等的 q 有多少个?

题解:标准解法是 KMP,我这双重循环强行水了过去..

#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
int a[N],b[N];
int main()
{
    int tcase,t=1;
    scanf("%d",&tcase);
    while(tcase--){
        int n,m,k;
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=m;i++){
            scanf("%d",&b[i]);
        }
        if(m>n) {
            printf("Case #%d: %d\n",t++,0);
            continue;
        }
        int ans = 0;
        for(int i=1;i<=n;i++){
            if(a[i]!=b[1]) continue;
            int idx = 1;
            for(int j=i;j<=n;j+=k){
                if(a[j]==b[idx]){
                    idx++;
                }
                else break;
                if(idx==m+1){
                    ans++;break;
                }
            }
        }
        printf("Case #%d: %d\n",t++,ans);
    }
    return 0;
}

 

 KMP解法

#include <iostream>
#include <cstring>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;

const int N = 1000002;
int Next[N];
int A[N],S[N], T[N];
int slen, tlen;

void getNext()
{
    int j, k;
    j = 0; k = -1; Next[0] = -1;
    while(j < tlen)
        if(k == -1 || T[j] == T[k])
            Next[++j] = ++k;
        else
            k = Next[k];

}
/*
返回模式串在主串S中出现的次数
*/
int KMP_Count()
{
    int ans = 0;
    int i, j = 0;
    if(slen == 1 && tlen == 1)
    {
        if(S[0] == T[0])
            return 1;
        else
            return 0;
    }
    for(i = 0; i < slen; i++)
    {
        while(j > 0 && S[i] != T[j])
            j = Next[j];
        if(S[i] == T[j])
            j++;
        if(j == tlen)
        {
            ans++;
            j = Next[j];
        }
    }
    return ans;
}
int main()
{

    int tcase,t=1;
    scanf("%d",&tcase);
    while(tcase--)
    {
        int n,k;
        scanf("%d%d%d",&n,&tlen,&k);

        memset(T,0,sizeof(T));
        for(int i=0;i<n;i++) scanf("%d",&A[i]);
        for(int i=0;i<tlen;i++) scanf("%d",&T[i]);
        int ans = 0;
        getNext();
        for(int i=0;i<k;i++){ ///枚举起点
            slen = 0;
            for(int j = i;i+(tlen-1)*k<n&&j<n;j+=k){
                S[slen++] = A[j];
            }
            if(slen<tlen) continue;
            /*for(int j=0;j<slen;j++){
                printf("%d ",S[j]);
            }*/
            ans+=KMP_Count();
        }
        printf("Case #%d: %d\n",t++,ans);
    }
    return 0;
}