Sumsets
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题目大意:给出一个数,求出和等于给出数的不同集合数,集合里的数只能是2的k次幂。
解题思路:定义dp[i]为数字i的最多集合数,当i为奇数时,dp[i] = dp[i - 1];当i为偶数时,dp[i - 1]中的每个集合加1得到dp[i]中所有含1的集合,而dp[i / 2]乘2得到dp[i]中只有偶数的集合。
代码如下:
#include <algorithm>
#include <cctype>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#define EPS 1e-6
#define INF INT_MAX / 10
#define LL long long
#define MOD 1000000000
#define PI acos(-1.0)
using namespace std;
const int maxn = 1000005;
LL dp[maxn];
int main()
{
int n;
scanf("%d",&n);
dp[1] = 1;
for(int i = 2;i <= n;i++){
if(i % 2){
dp[i] = dp[i - 1] % MOD;
}
else{
dp[i] = dp[i - 1] % MOD + dp[i >> 1] % MOD;
}
}
printf("%lld\n",dp[n]);
return 0;
}