POJ 2229 Sumsets (DP or math)

Description

Farmer John commanded his cows to search forDifferent sets of numbers that sum to a given number. The cows use is numbers that is an integer power of2. Here is the possible sets of numbers that sum to7: 1)1+1+1+1+1+1+1 2)1+1+1+1+1+2 3)1+1+1+2+2 4)1+1+1+4 5)1+2+2+2 6)1+2+4Help FJ count all possible representations forA given integer N (1<= N <=1, the, the).

Input

A single line with a single integer, N.

Output

 as  This 9 digits (inbase representation).

Sample Input

7

Sample Output

6

Source

Usaco 2005 January Silver

If I is an odd number, there must be a 1, the f[i-1] in each case plus a 1 to get fi, so f[i]=f[i-1]

If I is an even number, if there is 1, at least two, then f[i-2] each case plus two 1, get I, if not 1, the decomposition of each item in addition to 2, then get F[I/2]

So F[I]=F[I-2]+F[I/2]

Since just output the last 9 digits, don't forget modulo 1000000000

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5#include <stdlib.h>6#include <cmath>7 using namespacestd;8 #defineN 10100069 #defineMOD 1000000000Ten intN; One intDp[n]; A voidinit () { -      -dp[1]=1; thedp[2]=2; -      for(intI=3; i<n;i++){ -         if(i&1){ -dp[i]=dp[i-1]; +         } -         Else{ +dp[i]=dp[i-1]+dp[i/2]; Adp[i]%=MOD; at         } -     } - } - intMain () - { - init (); in      while(SCANF ("%d", &n) = =1){ -          toprintf"%d\n", Dp[n]); +      -     } the     return 0; *}

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POJ 2229 Sumsets (DP or math)