# [BZOJ1677][Usaco2005 Jan]Sumsets 求和

## 1677: [Usaco2005 Jan]Sumsets 求和

Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 1031  Solved: 603 [Submit][Status][Discuss]

## 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).

一个整数N.

7

## Sample Output

6

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
DP
#include <cstdio>
const int maxn = 1000000 + 10, mod = 1000000000;
int dp[maxn];
int main(){
int n;
scanf("%d", &n);
dp[1] = 1;
for(int i = 2; i <= n; i++)
if(i & 1) dp[i] = dp[i - 1];
else{
dp[i] = dp[i - 1] + dp[i >> 1];
if(dp[i] >= mod) dp[i] -= mod;
}
printf("%d\n", dp[n]);
return 0;
}

posted @ 2017-09-06 21:37  Elder_Giang  阅读(86)  评论(0编辑  收藏  举报