Sumsets
Time Limit: 2000MS | Memory Limit: 200000K | |
Total Submissions: 23002 | Accepted: 8810 |
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
找规律
递推 打表
当i为奇数时 a[i]=a[i-1];
当i为偶数时 a[i]=a[i/2]+a[i-2];
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#define ll long long
using namespace std;
ll a[1000002];
int main()
{
memset(a,0,sizeof(a));
a[1]=1;a[2]=2;a[3]=2;
a[4]=4;a[5]=4;
for(int i=6;i<=1000000;i++)
{
if(i%2==0)
a[i]=(a[i/2]+a[i-2])%1000000000;//不超过九个数字 所以对1000000000取模
else a[i]=a[i-1]%1000000000;
}
ll n;
cin>>n;
cout<<a[n]<<endl;
return 0;
}