HDU2709 Sumsets

Sumsets

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2484    Accepted Submission(s): 1001

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

当i为奇数的时候 a[i]=a[i-1];相当于在前一个数的加数后面都加上一；

当i为偶数的时候啊a[i]=a[i-2]+a[i>>1]：

1）加数里含有1，则一定含有两个1，相当于a[i-2]的加数里加上两个一，a[i-2]种；

2）加数里不含1，则相当于i/2 的加数都乘以2，a[i>>2]种

因此 a[i]=a[i-2]+a[i>>1];

#include <stdio.h>int a[1000005] = {1,1,2};int main(){int n;for(int i = 3; i <= 1000000; i ++){if(i % 2 == 0)a[i] = (a[i - 1] + a[i / 2]) % 1000000000;elsea[i] = a[i - 1];}while(~scanf("%d",&n)){printf("%d\n",a[n]);}return 0;}