【POJ】2229 Sumsets(递推)

Sumsets
Time Limit: 2000MS   Memory Limit: 200000K
Total Submissions: 20315   Accepted: 7930

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

7

Sample Output

6

Source

 
 
 
------------------------------------------------------------------------------------------------
分析:按照背包递推即可。
 
#include <cstdio>
#include <algorithm>
using namespace std;
int w[25],dp[1000005];
int main()
{
    int a=1,n;
    scanf("%d",&n);
    for(int i=0;i<=25;i++)
    {
        w[i]=a;
        a*=2;
    }
    dp[0]=1;
    for(int i=0;i<=25;i++)
    for(int j=w[i];j<=1000000;j++)
    {
        dp[j]+=dp[j-w[i]];
        dp[j]%=1000000000;
    }
    printf("%d",dp[n]%1000000000);
    return 0;
}

 

posted @ 2017-10-29 13:34  noble_(noblex)  阅读(188)  评论(0编辑  收藏  举报
/* */