HDU - 5918 Sequence I

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+(m1)p aq,aq+p,aq+2p,⋯,aq+(m−1)p where  q+(m1)pn q+(m−1)p≤n and  q1 q≥1.
Input
The first line contains only one integer  T100 T≤100, which indicates the number of test cases. 

Each test case contains three lines. 

The first line contains three space-separated integers  1n106,1m106 1≤n≤106,1≤m≤106 and  1p106 1≤p≤106

The second line contains n integers  a1,a2,,an(1ai109) a1,a2,⋯,an(1≤ai≤109)

the third line contains m integers  b1,b2,,bm(1bi109) 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

有规律的暴力,直接求答案:

#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: %d\n",M - N, sum);
	}
	return 0;
}


HDU1160是一个经典的动态规划问题,题目描述如下: 给定一组老鼠的体重和速度,要求找出最长的老鼠序列,使得序列中每只老鼠的体重和速度都严格递增。 我们可以使用动态规划来解决这个问题。具体步骤如下: 1. 将老鼠按体重或速度排序。 2. 使用动态规划数组`dp`,其中`dp[i]`表示以第`i`只老鼠结尾的最长序列长度。 3. 使用一个数组`pre`来记录每个位置的前驱老鼠,以便最后可以重建序列。 以下是使用Java实现的代码: ```java import java.util.*; class Mouse implements Comparable<Mouse> { int weight, speed, index; Mouse(int weight, int speed, int index) { this.weight = weight; this.speed = speed; this.index = index; } public int compareTo(Mouse other) { if (this.weight != other.weight) { return this.weight - other.weight; } else { return this.speed - other.speed; } } } public class Main { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); List<Mouse> mice = new ArrayList<>(); while (scanner.hasNext()) { int weight = scanner.nextInt(); int speed = scanner.nextInt(); mice.add(new Mouse(weight, speed, mice.size())); } scanner.close(); Collections.sort(mice); int n = mice.size(); int[] dp = new int[n]; int[] pre = new int[n]; int maxLen = 1, endIndex = 0; Arrays.fill(pre, -1); Arrays.fill(dp, 1); for (int i = 1; i < n; i++) { for (int j = 0; j < i; j++) { if (mice.get(i).weight > mice.get(j).weight && mice.get(i).speed > mice.get(j).speed) { if (dp[j] + 1 > dp[i]) { dp[i] = dp[j] + 1; pre[i] = j; } } } if (dp[i] > maxLen) { maxLen = dp[i]; endIndex = i; } } List<Integer> sequence = new ArrayList<>(); while (endIndex != -1) { sequence.add(mice.get(endIndex).index + 1); endIndex = pre[endIndex]; } System.out.println(maxLen); for (int i = sequence.size() - 1; i >= 0; i--) { System.out.println(sequence.get(i)); } } } ```
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