POJ 2229

Sumsets

Time Limit: 2000MS Memory Limit: 200000KTotal Submissions: 20111 Accepted: 7859

Description

Farmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers that are an integer power of 2. Here are the possible sets of numbers that sum to 7: 

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+4 

Help FJ count all possible representations for a given integer N (1 <= N <= 1,000,000). 

Input

A single line with a single integer, N.

Output

The number of ways to represent N as the indicated sum. Due to the potential huge size of this number, print only last 9 digits (in base 10 representation).

Sample Input

7

Sample Output

6

题意： 给你一个数n 问你可以多少种 2的某次方 相加表示。

简单题： 写出前几个数就可以推出dp公式 分奇偶数。

代码： 

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define N 1000005#define mod 1000000000using namespace std;long long dp[N];void init(){dp[1]=1;for(int i=2;i<N;i++){if(i%2) dp[i]=dp[i-1]%mod;else dp[i]=(dp[i-1]+dp[i/2])%mod;}return ;}int main(){int n;init();scanf("%d",&n);printf("%lld\n",dp[n]);return 0;}