Poj2229--Sumsets（递推）

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 15052		Accepted: 6001

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

Source

USACO 2005 January Silver

 

题目很有意思， 没想到是个递推， 脑子不够用。

#include <cmath>
#include <cstdio> 
const int  MOD = 1e9;
const int N = 1000001;
int num[N];
void Sieve(){
    num[1] = 1;
    num[2] = 2;
    num[3] = 2;
    for(int i = 4; i <= N; i++){
        if(i & 1)
            num[i] = num[i-1];
        else
            num[i] = (num[i-2]%MOD + num[i/2]%MOD)%MOD;
    }
}
int main(){
    Sieve();
    int n;
    while(scanf("%d", &n) != EOF)
        printf("%d
", num[n]);
    return 0;
}