link:http://acm.hdu.edu.cn/showproblem.php?pid=2709
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
整數遞推,這個題從1開始,結果依次爲 1 2 2 4 4 6 6 10 10 14 14 20 20 26 26 36 36 46 46 60 60 74
可以看出當這個數是奇數x時,a[x]=a[x-1]
偶數時,a[x]=a[x-2]+a[x/2]
因爲題目限制輸出後九位,所以都要對1e9取餘
AC代碼:
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int N=1000100;
int mod=1000000000;
int a[N];
void fun()
{
int i;
a[1]=1;
a[2]=2;
for(i=3;i<N;i++)
{
if(i%2==1)
a[i]=a[i-1]%mod;
else
a[i]=(a[i-2]+a[i/2])%mod;
}
}
int main()
{
int n;
fun();
while(~scanf("%d",&n))
printf("%d\n",a[n]);
return 0;
}