S - Sumsets

最新推荐文章于 2019-10-30 09:58:15 发布

原创最新推荐文章于 2019-10-30 09:58:15 发布 · 139 阅读

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博客围绕Farmer John让奶牛寻找和为给定整数N的不同数字集合展开，这些数字是2的整数次幂。给出了和为7的所有可能集合示例，还给出输入输出要求。题解指出，当N为奇数时，dp[n]=dp[n - 1]；当N为偶数时，dp[n]=dp[n - 1]+dp[n/2]。

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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

Sample Output

题解：当n为奇数时，dp[n] = dp[n-1],因为一定包含一个1，所以dp[n-1]的每一种再加1都一一对应着dp[n];当n为偶数时，要么至少含2个1，或者不含1，不含1时，方法数就是取dp[n/2]的每一种物品乘以2价值的物品。所以dp[n] = dp[n-1] + dp[n/2]。

#include<stdio.h>
#include<string.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n;
int s[1000010];
int main()
{
    while(~scanf("%d",&n))
    {
        s[0]=0;
        s[1]=1;
        s[2]=2;
        for(int i=3;i<=n;i++)
        {
            if(i%2==1)
                s[i]=s[i-1];
            else
            s[i]=(s[i-1]+s[i/2])%1000000000;
        }
        printf("%d\n",s[n]);
    }
}