POJ2229 Sumsets 【递归】

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 13210		Accepted: 5300

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

field=source&key=USACO+2005+January+Silver" style="text-decoration:none">USACO 2005 January Silver

做这题时有些被曾经的经验束缚了，看完题第一反应是母函数，然后上模板。然后输入1000000等结果，等啊等就是不出结果。。

。參考了大牛的解题报告：Click

#include <stdio.h>
#include <string.h>

#define maxn 1000002
int dp[maxn];

int main() {
    int i, n;
    scanf("%d", &n);
    dp[1] = 1;
    for(i = 2; i <= n; ++i) {
        if(i & 1) dp[i] = dp[i-1];
        else dp[i] = dp[i-1] + dp[i/2];
        dp[i] %= 1000000000;
    }
    printf("%d\n", dp[n]);
    return 0;
}

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