牛客网 2018校招真题 网易 小易喜欢的数列

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Description

牛客网 2018校招真题 小易喜欢的数列

Solving Ideas

动态规划：

State：
dp[i][j]: 表示长度为i且以数字j为结尾的合法数列的数量

Initial State:
dp[1][m] = 1; (1<=m<=k)

State Transition:
dp[i][j] += dp[i - 1][m] (1<=m<=k,且(m,j)是一个合法的数列) (i > 1)

为了降低求dp[i][j]的时间复杂度，可以先计算全部可能的排列数，不管(m,j)是否为合法数列，然后再计算非法的排列数invalid，则dp[i][j] = sum - invalid

Time complexity : $O(n*k*log(k))$
Space complexity : $O(n*k)$

Solution

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/**
 * Dynamic Programming
 *
 * State:
 *   dp[i][j]: 表示长度为i且以数字j为结尾的合法数列的数量
 *
 * Initial State:
 *   dp[1][m] = 1; (1<=m<=k)
 *
 * State Transition:
 *   dp[i][j] += dp[i - 1][m] (1<=m<=k,且(m,j)是一个合法的数列) (i > 1)
 *
 * @author wylu
 */
public class Main {
    static final int mod = 1000000007;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] strs = br.readLine().split(" ");
        int n = Integer.parseInt(strs[0]), k = Integer.parseInt(strs[1]);

        int[][] dp = new int[n + 1][k + 1];

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= k; j++){
                dp[i][j] = 1;
            }
        }

        for (int i = 2; i <= n; i++) {
            int sum = 0; //全部可能的排列数
            for (int j = 1; j <= k; j++) {
                sum = (sum + dp[i - 1][j]) % mod;
            }

            for (int j = 1; j <= k; j++) {
                int invalid = 0; //非法的排列数
                for (int m = 2 * j; m <= k; m += j) {
                    invalid = (invalid + dp[i - 1][m]) % mod;
                }
                //加模数避免(sum-invalid)为负的情况
                dp[i][j] = (sum - invalid + mod) % mod;
            }
        }

        int res = 0;
        for (int i = 1; i <= k; i++) {
            res = (res + dp[n][i]) % mod;
        }
        System.out.println(res);
    }
}