[BZOJ1677][Usaco2005 Jan]Sumsets 求和

1677: [Usaco2005 Jan]Sumsets 求和

Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 1031  Solved: 603 [Submit][Status][Discuss]

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).

给出一个N(1≤N≤10^6)，使用一些2的若干次幂的数相加来求之．问有多少种方法

Input

   一个整数N.

Output

方法数．这个数可能很大，请输出其在十进制下的最后9位．

Sample Input

7

Sample Output

6

有以下六种方式
    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

DP

#include <cstdio>
const int maxn = 1000000 + 10, mod = 1000000000;
int dp[maxn];
int main(){
    int n;
    scanf("%d", &n);
    dp[1] = 1;
    for(int i = 2; i <= n; i++)
        if(i & 1) dp[i] = dp[i - 1];
        else{
            dp[i] = dp[i - 1] + dp[i >> 1];
            if(dp[i] >= mod) dp[i] -= mod;
        }
    printf("%d
", dp[n]);
    return 0;
}