hdu2709(DP)

Sumsets

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1294    Accepted Submission(s): 495

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

7

 

Sample Output

6

 

Source

USACO 2005 January Silver

 

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            本题要求所给数的加数组合，其中加数为2的整数幂。数据量很大，所有暗示有O(n)的算法。

          dp[1]=1,dp[2]=2.

         当n为奇数时，必含有一个加数1，此时dp[n]=dp[n-1];

         当n为偶数时，要么不含1，至少含有两个1，此时dp[n]=dp[n-2]+dp[n/2],不含1与n/2时等价，至少含2个1与dp[n-2]等价。

#include<iostream>using namespace std;const int MAXN=1000000+100;const int MOD=1000000000;__int64 dp[MAXN];void DP(){memset(dp,0,sizeof(dp));dp[1]=1;dp[2]=2;for(int i=3;i<=1000001;i++){if(0==i%2)dp[i]=(dp[i-2]+dp[i/2])%MOD;else dp[i]=dp[i-1];}}int main(){int n;DP();while(~scanf("%d",&n))printf("%I64d\n",dp[n]);return 0;}