Sequence I
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1013 Accepted Submission(s): 393
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
题意：
题解：
kmp
1 /******************************
2 code by drizzle
3 blog: www.cnblogs.com/hsd-/
4 ^ ^ ^ ^
5 O O
6 ******************************/
7 #include<bits/stdc++.h>
8 #include<map>
9 #include<set>
10 #include<cmath>
11 #include<queue>
12 #include<bitset>
13 #include<math.h>
14 #include<vector>
15 #include<string>
16 #include<stdio.h>
17 #include<cstring>
18 #include<iostream>
19 #include<algorithm>
20 #pragma comment(linker, "/STACK:102400000,102400000")
21 using namespace std;
22 #define A first
23 #define B second
24 const int mod=1000000007;
25 const int MOD1=1000000007;
26 const int MOD2=1000000009;
27 const double EPS=0.00000001;
28 //typedef long long ll;
29 typedef __int64 ll;
30 const ll MOD=1000000007;
31 const int INF=1000000010;
32 const ll MAX=1ll<<55;
33 const double eps=1e-5;
34 const double inf=~0u>>1;
35 const double pi=acos(-1.0);
36 typedef double db;
37 typedef unsigned int uint;
38 typedef unsigned long long ull;
39 int f[2000100];
40 void get(int *p,int m)
41 {
42 int j=0;
43 f[0]=f[1]=0;
44 for(int i=1;i<m;i++)
45 {
46 j=f[i];
47 while(j&&p[j]!=p[i]) j=f[j];
48 if(p[i]==p[j]) f[i+1]=j+1;
49 else f[i+1]=0;
50 }
51 }
52 int kmp(int *s,int *p,int n,int m)
53 {
54 int num=0;
55 int j=0;
56 for(int i=0;i<n;i++)
57 {
58 while(j&&p[j]!=s[i]) j=f[j];
59 if(s[i]==p[j]) j++;
60 if(j==m) num++;
61 }
62 return num;
63 }
64 int s[2000100],p[2000100],t[2000100];
65 int n,m,q;
66 int main()
67 {
68 int T;
69 scanf("%d",&T);
70 for(int k=1;k<=T;k++)
71 {
72 scanf("%d %d %d",&n,&m,&q);
73 memset(t,0,sizeof(t));
74 memset(p,0,sizeof(p));
75 for(int i=0;i<n;i++)
76 scanf("%d",&t[i]);
77 for(int i=0;i<m;i++)
78 scanf("%d",&p[i]);
79 get(p,m);
80 int ans=0;
81 for(int i=0;i<q;i++)
82 {
83 int num=0;
84 for(int j=i;j<n&&i+(m-1)*q<n;j+=q)
85 s[num++]=t[j];
86 ans+=kmp(s,p,num,m);
87 }
88 printf("Case #%d: %d\n",k,ans);
89 }
90 return 0;
91 }
92 /*
93 KMP 处理
94 *
模拟
1 /******************************
2 code by drizzle
3 blog: www.cnblogs.com/hsd-/
4 ^ ^ ^ ^
5 O O
6 ******************************/
7 #include<bits/stdc++.h>
8 #include<map>
9 #include<set>
10 #include<cmath>
11 #include<queue>
12 #include<bitset>
13 #include<math.h>
14 #include<vector>
15 #include<string>
16 #include<stdio.h>
17 #include<cstring>
18 #include<iostream>
19 #include<algorithm>
20 #pragma comment(linker, "/STACK:102400000,102400000")
21 using namespace std;
22 #define A first
23 #define B second
24 const int mod=1000000007;
25 const int MOD1=1000000007;
26 const int MOD2=1000000009;
27 const double EPS=0.00000001;
28 //typedef long long ll;
29 typedef __int64 ll;
30 const ll MOD=1000000007;
31 const int INF=1000000010;
32 const ll MAX=1ll<<55;
33 const double eps=1e-5;
34 const double inf=~0u>>1;
35 const double pi=acos(-1.0);
36 typedef double db;
37 typedef unsigned int uint;
38 typedef unsigned long long ull;
39 int t;
40 int a[1000006];
41 int b[1000006];
42 int d[1000006];
43 map<int,int> mp;
44 vector<int > ve[1000006];
45 int n,m,p;
46 int main()
47 {
48 while(scanf("%d",&t)!=EOF)
49 {
50 for(int l=1; l<=t; l++)
51 {
52 mp.clear();
53 scanf("%d %d %d",&n,&m,&p);
54 for(int i=1; i<=n; i++)
55 scanf("%d",&a[i]);
56 int jishu=1;
57 for(int i=1; i<=m; i++ )
58 {
59 ve[i].clear();
60 scanf("%d",&b[i]);
61 if(mp[b[i]]==0)
62 {
63 mp[b[i]]=jishu;
64 d[jishu]=i;
65 jishu++;
66 }
67 }
68 int minx=10000000;
69 int what=0;
70 for(int i=1; i<=n; i++)
71 {
72 if(mp[a[i]])
73 {
74 ve[mp[a[i]]].push_back(i);
75 }
76 }
77 for(int i=1; i<jishu; i++)
78 {
79 if(minx>ve[i].size())
80 {
81 minx=ve[i].size();
82 what=i;
83 }
84 }
85 int sum=0;
86 for(int i=0; i<ve[what].size(); i++)
87 {
88 int st=ve[what][i]-(d[mp[a[ve[what][i]]]]-1)*p;
89 int ed=ve[what][i]+(m-d[mp[a[ve[what][i]]]])*p;
90 int zha=1;
91 int flag=0;
92 if(st<1||ed>n)
93 flag=1;
94 if(flag==0)
95 {
96 for(int j=st; j<=ed; j+=p)
97 {
98 if(a[j]!=b[zha++])
99 {
100 flag=1;
101 break;
102 }
103 }
104 }
105 if(flag==0)
106 sum++;
107 }
108 printf("Case #%d: %d\n",l,sum);
109 }
110 }
111 return 0;
112 }
时间： 2024-10-02 09:18:02