POJ2229 Sumsets 【递推】

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

USACO 2005 January Silver

做这题时有些被以前的经验束缚了，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。。。参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
);
return 0;
}