Sumsets

最新推荐文章于 2022-12-17 13:35:59 发布

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Sumsets

Time Limit : 6000/2000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 16 Accepted Submission(s) : 12

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Problem 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

Sample Output

分析：

本题是一道规律题，分两种情况，一种是奇数时，很明显此时，但是偶数是找不到，网上看了大神的代码，看不懂，待日后理解。

代码：

#include<stdio.h>
__int64 a[1000010];
void f()
{
	int i,j,k;
	a[1]=1;
	a[2]=2;
	for(i=3;i<1000010;i++)
	{
		if(i%2)
        a[i]=a[i-1];
		else a[i]=a[i-2]+a[i/2];
		a[i]%=1000000000;		
	}
}
int main()
{
	__int64 n;
	f(); //dabiao
	while(~(scanf("%I64d",&n)))
	{
		printf("%I64d\n",a[n]);
		
	}
	return 0;
}

方法二：

动态规划：网上看的，不懂，粘下来慢慢理解。

状态：
d[i][j]表示前i个二的幂数凑成数j的方法数
空间可以降维到d[j]
状态转移方程：
d[j]=d[j]+d[j-c[i]]
c[i]=2^i
边界：
d[0]=1
代码：
#include<cstdio>


int d[1000005],c[25],n,i,j;
int main()
{
	scanf("%d",&n);
	c[0]=d[0]=1;
	for(i=1;i<=20;i++)
		c[i]=c[i-1]<<1;
	for(i=0;i<=20&&c[i]<=n;i++)
		for(j=c[i];j<=n;j++)
			d[j]=(d[j]+d[j-c[i]])%1000000000;
	printf("%d/n",d[n]);		
}