Sumsets
Time Limit: 2000MS | Memory Limit: 200000K | |
Total Submissions: 15067 | Accepted: 6007 |
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
思路:对于奇数来说,每个式子+1,所以式子个数不变;对于偶数来说,比n-2多2,除了n-1的那些式子+1,还有以2为基数的组成的式子,即s[n/2]个。。。。
AC-code:
#include<cstdio> int s[1000005]; int main() { int n,i; scanf("%d",&n); s[1]=1; s[2]=2; for(i=3;i<=n;i++) { if(i%2) s[i]=s[i-1]; else { s[i]=s[i-2]+s[i/2]; s[i]%=1000000000; } } printf("%d\n",s[n]); return 0; }