POJ 2229 Sumsets

Sumsets

Time Limit: 2000MS Memory Limit: 200000K

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

思路：

本题相当于分成，有1和没1的情况，有1的话等价于i-1，没1的话说明都是2的幂次，方案数等价于全部缩小一倍，比如7，如果拆开的里面有1，那么相当于最前面加了个1，后面就是6的情况，如果没有1，也就是都是2.4.8这类的组合，问等于多少，那就可以看成2,4,8缩小一倍以后的总方案数，相当于比如问只能用2.4.8.16…组合32的方案数，那其实就等于用1.2.4.8.16…组合16的个数，而这个是个子问题

#include <iostream>
using namespace std;
const int maxn = 1e+9;
int dp[1000010];
int main() {
	int n;
	cin >> n;
	dp[1] = 1;
	for (int i = 2; i <= n; i++) {
		if (i % 2 == 1) dp[i] = dp[i - 1] % maxn;
		else dp[i] = (dp[i - 1] + dp[i / 2]) % maxn;
	}
	cout << dp[n] << endl;
	return 0;
}