Sumsets

Time Limit : 6000/2000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 45 Accepted Submission(s) : 20

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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

Sample Output

6
题意 给定一个n，n又2的幂次方相加得到，问有多少中相加的方式
分析 当n为奇数的时候 就是再前一个的基础上加上1，a[n]=a[n-1]
　　当n为偶数的时候：
　　　　如果加数里含有1，则一定至少有2个1，就是对n-2后面+1+1，就是a[n-2]
　　　　如果加数里面没有1，即对n/2的每一个加数乘以2，总类数为a[n/2]
　　所以n为偶数时的总类数为a[n]=a[n-2]+a[n/2]

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<string>
 4 #include<cmath>
 5 #include<iostream>
 6 #include<algorithm>
 7 #include<map>
 8 #include<set>
 9 #include<queue>
10 #include<vector>
11 using namespace std;
12 int a[1000000+5];
13 int main()
14 {
15     int n;
16     a[1]=1,a[2]=2;
17     for(int i=3;i<=1000000;i++)
18     {
19         if(i%2)
20         {
21             a[i]=a[i-1];
22         }
23         else
24         {
25             a[i]=a[i-2]+a[i/2];
26             a[i]%=1000000000;
27         }
28     }
29     while(~scanf("%d",&n))
30     {
31         printf("%d\n",a[n]);
32     }
33     return 0; 
34 }

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