目录

 

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1 问题描述

Problem Description

　　We call a number interesting, if and only if:

　　1. Its digits consists of only 0, 1, 2 and 3, and all these digits occurred at least once.

　　2. Inside this number, all 0s occur before any 1s, and all 2s occur before any 3s.

　　Therefore, the smallest interesting number according to our definition is 2013. There are two more interseting number of 4 digits: 2031 and 2301.

　　Your task is to calculate the number of interesting numbers of exactly n digits. As the answer might be very large, you only need to output the answer modulo 1000000007.

Input Format

　　The input has one line consisting of one positive integer n (4 ≤ n ≤ 10^15).

Output Format

　　The output has just one line, containing the number of interesting numbers of exactly n digits, modulo 1000000007.

Input Sample

　　4

Output Sample

　　3

 

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2 解决方案

本题主要考查数学组合数推理化简，具体思考过程如下：

引用自文末：

 

推导过程，在草稿纸上推导了一下：

本文下面代码结果运行为90分，代码仅供参考，文末参考资料1中代码运行结果为100分，可以参考一下哦。

具体代码如下：

import java.util.Scanner;public class Main {    public final static long p = 1000000007L;    //求取a的b次方取余p的值    public long getPowMod(long a, long b) {        long temp = a, result = 1;        while(b != 0) {            if((b & 1) == 1)                result = result * temp % p;            temp = temp * temp % p;            b >>= 1;        }        return result;    }        public void printResult(long n) {        if(n == 4) {            System.out.println(3);            return;        }        long m = n - 1;        long result = getPowMod(2, m - 2);        m = m % p;        result = result * (m * m % p - 3 * m % p) % p + m;        result %= p;        System.out.println(result);        return;    }        public static void main(String[] args) {        Main test = new Main();        Scanner in = new Scanner(System.in);        long n = in.nextLong();        test.printResult(n);            }}

 

 

 

参考资料：

1.