HDU - 5918 Sequence I

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

Input

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

复制代码

1
2
3
4
5
6
7
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

复制代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
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27
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31
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33
34
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51
52
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55
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58
59
60
61
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67
68
Case #1: 2
Case #2: 1


有规律的暴力，直接求答案：
复制代码#include <stdio.h>
const int maxn = 1e6+5;
int a[maxn];
int b[maxn];
int main() {
	int N;
	int n , m , p;
	scanf("%d",&N);
	int M = N;
	while(N--) {
		int sum = 0 , flag;
		scanf("%d %d %d",&n, &m, &p);
		for(int i = 0; i < n; i++)
		scanf("%d",&a[i]);
		for(int i = 0; i < m; i++)
		scanf("%d",&b[i]);
		for(int j = 0; j <= (n - m*p + p - 1); j++) {
			if(a[j] == b[0]) {
				flag = 0;
				for(int k = 1; k < m; k++) {
					if(b[k] != a[j + k*p]) {
						flag = 1;
						break;
					}
				}	
				if(flag == 0)
				sum++;
			}
		}
		printf("Case #%d: %dn",M - N, sum);
	}
	return 0;
}#include <stdio.h>
const int maxn = 1e6+5;
int a[maxn];
int b[maxn];
int main() {
	int N;
	int n , m , p;
	scanf("%d",&N);
	int M = N;
	while(N--) {
		int sum = 0 , flag;
		scanf("%d %d %d",&n, &m, &p);
		for(int i = 0; i < n; i++)
		scanf("%d",&a[i]);
		for(int i = 0; i < m; i++)
		scanf("%d",&b[i]);
		for(int j = 0; j <= (n - m*p + p - 1); j++) {
			if(a[j] == b[0]) {
				flag = 0;
				for(int k = 1; k < m; k++) {
					if(b[k] != a[j + k*p]) {
						flag = 1;
						break;
					}
				}	
				if(flag == 0)
				sum++;
			}
		}
		printf("Case #%d: %dn",M - N, sum);
	}
	return 0;
}