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