Sumsets（POJ 2229 DP）
Sumsets
| Time Limit: 2000MS | Memory Limit: 200000K | |
| Total Submissions: 15293 | Accepted: 6073 |
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
分两种情况讨论：
当n为奇数时，划分中肯定存在1，那么dp[i]=dp[i-1];
当n为偶数时，可以把划分情况分为含有1和没有1两种情况
含有1，那么i-1为奇数,dp[i-1];
不含有1,那么划分中都是偶数，相当于dp[i>>1]的每种划分的两倍；
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 int M=1000000000; 6 unsigned long long dp[1000005]; 7 int main() 8 { 9 unsigned long long n,i,j; 10 scanf("%lld",&n); 11 dp[1]=1; 12 for(i=2;i<=n;i++) 13 dp[i] = (i%2 ?dp[i-1]:(dp[i-1]+dp[i>>1]))%M; 14 printf("%lld\n",dp[n]); 15 return 0; 16 }
