51Nod 1048 1383 整数分解为2的幂

任何正整数都能分解成2的幂，给定整数N，求N的此类划分方法的数量！

比如N = 7时，共有6种划分方法。

 

7=1+1+1+1+1+1+1

  =1+1+1+1+1+2

  =1+1+1+2+2

  =1+2+2+2

  =1+1+1+4

  =1+2+4

 

傻逼的高精度DP，我只选Java BigInteger。

 

import java.util.*;
import java.math.*;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        
        BigInteger N = in.nextBigInteger();
        
        BigInteger F[][] = new BigInteger[125][125];
        BigInteger G[][] = new BigInteger[125][125];
        
        for (int i = 0; i <= 120; ++i)
            for (int j = 0; j <= 120; ++j) {
                F[i][j] = BigInteger.ZERO;
                G[i][j] = BigInteger.ZERO;
            }
        
        F[0][0] = BigInteger.ONE;
        
        for (int i = 1; i <= 120; ++i) {
            F[i][i] = BigInteger.ONE;
            
            for (int j = 0; j < i; ++j)
                for (int k = 0; k <= j; ++k)
                    F[i][j] = F[i][j].add(F[i - 1][k].multiply(F[i - k - 1][j - k]));
        }
        
        int tot = 0;
        
        BigInteger TWO = BigInteger.ONE.add(BigInteger.ONE);
        
        for (int i = 0; i <= 120; ++i) {
            if (N.mod(TWO).toString().charAt(0) == '1') {
                if (++tot == 1) {
                    for (int j = 0; j <= i; ++j)
                        G[tot][j] = F[i][j];
                }
                else {
                    for (int j = 0; j <= i; ++j)
                        for (int k = 0; k <= j; ++k)
                            G[tot][j] = G[tot][j].add(G[tot - 1][k].multiply(F[i - k][j - k]));
                }
            }
            
            N = N.divide(TWO);
        }
        
        BigInteger ANS = BigInteger.ZERO;
        
        for (int i = 0; i <= 120; ++i)
            ANS = ANS.add(G[tot][i]);
        
        System.out.println(ANS.toString());
    }
}

 

@Author: YouSiki