【37.68%】【hdu 5918】Sequence I

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

Total Submission(s): 706 Accepted Submission(s): 266

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

【题解】



意思是让你在

a1,a1+p,a1+2p,a1+3p…

a2,a2+p,a2+2p,a2 + 3p..

a3,a3+p,a3+2p,a3 +3p

…

ap,ap+p,ap+2p,ap+3p..

(a右边的东西都是下标;)

这p个序列里面找b数组的匹配数目；

用vector类处理出这个p个数列就好。

剩下的用KMP算法解决。

找完一个匹配之后,j==m。

这个时候让j= f[j];

就能继续找匹配了。

记住就好。不然每次都想好烦。

#include <cstdio>
#include <iostream>
#include <vector>
const int MAXN = 2e6;
const int MAXM = 2e6;
using namespace std;
int p;
vector <int> a[MAXN];
int b[MAXM];
int f[MAXM],ans,n,m;
void input(int &r)
{
    int t = getchar();
    while (!isdigit(t)) t = getchar();
    r = 0;
    while (isdigit(t)) r = r * 10+t-'0', t = getchar();
}
int main()
{
    //freopen("F:\\rush.txt", "r", stdin);
    int t;
    input(t);
    for (int q = 1; q <= t; q++)
    {
        for (int i = 0; i <= 1000000; i++)//a[0]也要clear!
            a[i].clear();
        ans = 0;
        input(n); input(m); input(p);
        for (int i = 1; i <= n; i++)
        {
            int x;
            input(x);
            a[(i%p)].push_back(x);
        }
        for (int j = 0; j <= m - 1; j++)
            input(b[j]);
        f[0] = 0; f[1] = 0;
        for (int i = 1; i <= m - 1; i++)//获取失配函数,b数组下表是从0开始的。
        {
            int j = f[i];
            while (j && b[j] != b[i]) j = f[j];
            f[i + 1] = b[j] == b[i] ? j + 1 : 0;
        }
        for (int i = 0; i <= p - 1; i++)//给p个数列找匹配数目
        {
            int j = 0, len = a[i].size();
            for (int k = 0; k <= len - 1; k++)
            {
                while (j && a[i][k] != b[j]) j = f[j];
                if (a[i][k] == b[j])
                    j++;
                if (j == m)
                {
                    ans++;
                    j = f[j];
                }
            }
        }
        printf("Case #%d: %d\n", q, ans);
    }
    return 0;
}