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;
}
