hdu2709Sumsets 找规律

Sumsets

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2623    Accepted Submission(s): 1056

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

规律：

当i为奇数的时候 dp[ i ]=dp[ i-1];

当i为偶数的时候啊dp[ i ]=dp[ i-1 ]+dp[ i / 2 ]：

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;typedef long long ll;ll dp[1000010];int main(){ll n,ans=2,k=4,sum=0,mod=1e9;dp[1]=1,dp[2]=dp[3]=2,dp[4]=dp[5]=4;for(ll i=6;i<=1000005;i++){if(i%2==1) dp[i]=dp[i-1];elsedp[i]=(dp[i-1]+dp[i/2])%mod;}while(scanf("%lld",&n)!=EOF){printf("%lld\n",dp[n]);}return 0;}