描述
http://poj.org/problem?id=2229
将一个数n分解为2的幂之和共有几种分法?
Sumsets
| Time Limit: 2000MS | Memory Limit: 200000K | |
| Total Submissions: 16207 | Accepted: 6405 |
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).
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
分析
对i讨论:
1.i是奇数:
分成的序列中必有1,所以可将i分为1+(i-1),所以f[i]=f[i-1];
2.i是偶数:
(1).分成的序列中有1:
同奇数,f[i]=f[i-1];
(2).分成的序列中没有1:
序列中的所有数都是2的倍数,那么任一种序列中的各个数/2,就得到了i/2的序列,那这种情况下,i的序列数就和i/2的序列数相同即f[i]=f[i/2];
综上:f[i]=f[i-1]+f[i/2];
1 #include<cstdio> 2 3 const int maxn=1000005,mod=1e9; 4 int n,f[maxn]; 5 6 int main() 7 { 8 #ifndef ONLINE_JUDGE 9 freopen("sum.in","r",stdin); 10 freopen("sum.out","w",stdout); 11 #endif 12 scanf("%d",&n); 13 f[1]=1; 14 for(int i=2;i<=n;i++) 15 { 16 if(i&1) f[i]=f[i-1]; 17 else f[i]=(f[i-1]+f[i/2])%mod; 18 } 19 printf("%d ",f[n]); 20 #ifndef ONLINE_JUDGE 21 fclose(stdin); 22 fclose(stdout); 23 #endif 24 return 0; 25 }