hdoj-2709-Sumsets【规律】

最新推荐文章于 2018-11-11 14:50:16 发布

原创最新推荐文章于 2018-11-11 14:50:16 发布 · 675 阅读

0 ·

CC 4.0 BY-SA版权

规律专栏收录该内容

3 篇文章

订阅专栏

Sumsets

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1869 Accepted Submission(s): 730

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

Source

USACO 2005 January Silver

Recommend

teddy | We have carefully selected several similar problems for you: 2544 2716 2710 2717 2715

之前用母函数做的，一直超时！！

//母函数 
#include<stdio.h> 
#include<string.h>
__int64 f[40]={1,2,4,8,16,32,64,128,256,512,1024,2048};
//,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864};
//,2147483648};
__int64 c1[1000001],c2[1000001];
__int64 N,i,j,k;
int main()
{
    
    while(~scanf("%I64d",&N))
    {  
        memset(c1,0,sizeof(c1));
        memset(c2,0,sizeof(c2));
        for(i=0;i<=N;++i) 
        c1[i]=1;
        for(i=1;f[i]<=N;++i)
        {
            for(j=0;j<=N;j++)
            {
                for(k=0;k<=N;k+=f[i])    
                c2[k+j]+=c1[j],c2[k+j]%=100000000;
            }
            for(j=0;j<=N;++j)
            c1[j]=c2[j],c2[j]=0;
            //printf("%d\n..",c1[N]);
        }
       printf("%d\n",c1[N])    ;
    }
    return 0;
}

#include<stdio.h>
#include<string.h>
int dp[1000010];
int main(){
	int n;	
	memset(dp,0,sizeof(dp));
	dp[0]=1;
	for(int i=1;i<=1000000;i*=2){
		for(int j=i;j<=1000000;++j){
			dp[j]+=dp[j-i];
			dp[j]%=1000000000;
		}
	}
	while(~scanf("%d",&n)){	
		printf("%d\n",dp[n]);
	}
	return 0;
}