Sumsets
| Time Limit: 2000MS | Memory Limit: 200000K | |
| Total Submissions: 13291 | Accepted: 5324 |
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).
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题目翻译：给出一个数字n，给出一个集合{1,2,4,8,16,32,64。。。。}，求n有多少种不同的由集合内的数字的加和情况。
解题思路：本以为是母函数，但是在网上搜了一下结果发现是递推，可以有前面推出后面的结果，原因是和2的指数关系有关。
#include<cstdio>
int dp[1000000+5]={0,1,2,2},i,n;
int main(){
while(scanf("%d",&n)==1){
for(i=4;i<=n+1;i+=2){
dp[i-1]=dp[i-2];
dp[i]=(dp[i-1]+dp[i>>1])%1000000000;
}
printf("%d\n",dp[n]);
}
return 0;
}
