hdu2709——Sumsets

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

数学题…用到递归，题目就是说将一个整数分解为2的幂相加（如4=1+1+2）共有多少种方法
思路是这样的，首先a[0]和a[1]都是1，从2开始，如果i是奇数，即i=1+(i-1)那a[i]肯定等于a[i-1]+1，而如果i是偶数，那么两种情况，如果分解的式子中存在1，那一定存在两个1，所以i=1+1+(i-2)，即a[i]=a[i-2]+2，而如果不存在1，那么说明分解的式子中全都是偶数，那么也就是整个式子可以除以2，即a[i]=a[i/2]

代码：

#include <cstdio>
#include <algorithm>
using namespace std;
long long arr[1000010];
int main(void){
    int n;
    arr[0]=1;
    arr[1]=1;
    for(int i=2;i<=1000000;i++){
        if(i%2==0){
            arr[i]=(arr[i-2]+arr[i/2])%1000000000;
        }
        else{
            arr[i]=arr[i-1]%1000000000;
        }
    }
    while(~scanf("%d",&n)){
        printf("%lld\n",arr[n]%1000000000);
    }
    return 0;
}

注意看清楚题意然后取模

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