project euler113

Posted 2020-10-10

project euler 113

对于1个数字，如果他不减或者不增称为bouncy number。统计1～10^100中的bouncy number

 

思路：分为两种计算

1.这个数中只出现了一个数字，那么对于i位的数字有9种，故总计有9 * 100种

2.这个数中至少出现两个数字，分别按位数计算不减的数和不增的数

以不减的数为例，枚举首数字j和尾数字k(j < k)，

j = a1 <= a2 <= a3 <= a4 <= a5 <= … <= am = k

令x1 = a2 - a1，x2 = a3 - a2, x3 = a4 - a3, …, xm - 1 = am - am-1。

x1 + x2 + x3 + … + xm-1 = k - j 解不定方程的解，C(k - j + m - 2, k - j)

不增的数同理

import java.util.*;
import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;

public class Main {
    BigInteger bi(int x) {
        return BigInteger.valueOf(x);
    }
    BigInteger comb(int n, int m) {
        BigInteger ret = bi(1);
        for (int i = n; i >= n - m + 1; -- i) ret = ret.multiply(bi(i));
        for (int i = 1; i <= m; ++ i) ret = ret.divide(bi(i));
        return ret;
    }
    void solve() {
        int n = 100;
        BigInteger ans = bi(9 * n);
        for (int i = 2; i <= n; ++ i) {
            for (int j = 1; j <= 8; ++ j)
                for (int k = j + 1; k <= 9; ++ k) {
                    ans = ans.add(comb(k - j + i - 2, k - j));
                }
        }
        for (int i = 2; i <= n; ++ i) {
            for (int j = 1; j <= 9; ++ j) 
                for (int k = 0; k <= j - 1; ++ k) {
                    ans = ans.add(comb(j - k + i - 2, j - k));
                }
        }
        System.out.println(ans);
     }
    FastScanner cin;
    PrintWriter out;
    void run() {
        cin = new FastScanner();
        out = new PrintWriter(System.out);
        solve();
        out.flush();
    }
    class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        boolean hasNext;
        
        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
            hasNext = true;
        }

//        public FastScanner(String s) {
//            try {
//                br = new BufferedReader(new FileReader(s));
//            } catch (FileNotFoundException e) {
//                // TODO Auto-generated catch block
//                e.printStackTrace();
//            }
//            hasNext = true;
//        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                    hasNext = false;
                    return "##";
                }
            }
            return st.nextToken();
        }
        String next = null;
        public boolean hasNext() {
            next = nextToken();
            return hasNext;
        }
        public int nextInt() {
            if (next == null) next = nextToken();
            String more = next;
            next = null;
            return Integer.parseInt(more);
        }
        public long nextLong() {
            return Long.parseLong(nextToken());
        }

        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
    public static void main(String[] args) {
        new Main().run();
    }
}