題干：

Mr. Frog has two sequences?a1,a2,?,ana1,a2,?,an?and?b1,b2,?,bmb1,b2,?,bm?and a number p. He wants to know the number of positions q such that sequence?b1,b2,?,bmb1,b2,?,bm?is exactly the sequence?aq,aq+p,aq+2p,?,aq+(m?1)paq,aq+p,aq+2p,?,aq+(m?1)p?where?q+(m?1)p≤nq+(m?1)p≤n?and?q≥1q≥1.

Input

The first line contains only one integer?T≤100T≤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≤1061≤n≤106,1≤m≤106?and?1≤p≤1061≤p≤106.?

The second line contains n integers?a1,a2,?,an(1≤ai≤109)a1,a2,?,an(1≤ai≤109).?

the third line contains m integers?b1,b2,?,bm(1≤bi≤109)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

題目大意：

? ? ?給兩個字符串a和b，大小分別是n和m，a數組以p為步長移動，b數組以1以步長移動，a中有多少個b串。（即數組匹配問題，只不過是a串的步長不是1）

? ? ?這題直接暴力o(nm)的復雜度可以過，也可以用KMP算法，一種專門處理字符串匹配問題的算法，處理長串的表現十分優秀，這里不做過多解釋。

AC代碼：

#include<bits/stdc++.h> using namespace std; int n,m,p,ans; int a[1000000 + 5],b[1000000 + 5]; int main() {int t,iCase = 0,flag;cin>>t;while(t--) {scanf("%d%d%d",&n,&m,&p);ans = 0;for(int i = 1; i<=n; i++) scanf("%d",&a[i]);for(int i = 1; i<=m; i++) scanf("%d",&b[i]);for(int i = 1; i+(m-1)*p<=n; i++) {flag = 1;for(int j = i; j<=i+(m-1)*p; j+=p) {if(a[j] != b[(j-i)/p+1]) {flag=0;break; }}if(flag) ans++;}printf("Case #%d: %d\n",++iCase,ans);}return 0 ; }

?

總結

以上是生活随笔為你收集整理的【HDU - 5918 】Sequence I （数组(字符串)匹配问题，可选KMP）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。