Sumsets
Time Limit: 2000MS | Memory Limit: 200000K | |
Total Submissions: 15052 | Accepted: 6001 |
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
题目很有意思, 没想到是个递推, 脑子不够用。
#include <cmath> #include <cstdio> const int MOD = 1e9; const int N = 1000001; int num[N]; void Sieve(){ num[1] = 1; num[2] = 2; num[3] = 2; for(int i = 4; i <= N; i++){ if(i & 1) num[i] = num[i-1]; else num[i] = (num[i-2]%MOD + num[i/2]%MOD)%MOD; } } int main(){ Sieve(); int n; while(scanf("%d", &n) != EOF) printf("%d ", num[n]); return 0; }