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HDU 5918 Sequence I KMP

2024-08-15 16:16:45   218人阅读

Sequence I



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,?,bmis 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
 

题意：

　　给你a,b两个序列

　　和一个p ，求有多少个 q恰好满足 b1,b2,b3....bm 就是 a[q],a[q+p],a[q+2p]......a[q+(m-1)p];

题解：

　　将a序列，每隔p位置分成一组，这样最多有p组，个数和是n

　　将每组和b序列跑KMP计算答案

#include<bits/stdc++.h>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")#define ls i<<1#define rs ls | 1#define mid ((ll+rr)>>1)#define pii pair<int,int>#define MP make_pairtypedef long long LL;const long long INF = 1e18;const double Pi = acos(-1.0);const int N = 1e6+10, M = 1e6, mod = 1e9+7, inf = 2e9;int T,n,m,p,s[N],t[N],ans = 0;vector<int > P[N];int nex[N];int main(){    int cas = 1;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&n,&m,&p);        for(int i = 1; i <= n; ++i) scanf("%d",&t[i]);        for(int i = 1; i <= m; ++i) scanf("%d",&s[i]);        for(int i = 0; i < p; ++i) P[i].clear();        memset(nex,0,sizeof(nex));        ans = 0;        int k = 0;        for(int i=2; i<=m; i++)        {            while(k>0&&s[k+1]!=s[i]) k = nex[k];            if(s[k+1]==s[i])k++;            nex[i] = k;        }        if(p == 1)        {            k = 0;            for(int i=1; i<=n; i++)            {                while(k>0&&s[k+1]!=t[i]) k = nex[k];                if(s[k+1]==t[i]) k++;                if(k==m) {                    k = nex[k];                    ans++;                }            }            printf("Case #%d: ",cas++);            printf("%d\n",ans);        }        else {            for(int i = 1; i <= n; ++i)                P[i % p].push_back(t[i]);            for(int i = 0; i < p; ++i) {                k = 0;                for(int j = 0; j < P[i].size(); ++j) {                    while(k>0&&s[k+1]!=P[i][j]) k = nex[k];                    if(s[k+1]==P[i][j]) k++;                    if(k==m) {                        k = nex[k];                        ans++;                    }                }            }            printf("Case #%d: ",cas++);            printf("%d\n",ans);        }    }}

 

HDU 5918 Sequence I KMP


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