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).
InputA single line with a single integer, N.OutputThe 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
题目大意就是将一个整数拆分为只有 1 或者 有 1 和偶数(2的次方数)组分的几个数的和
当于在f[n-1]的拆分结果上增加一个 1 即可;
当整数n为偶数时,分为两种情况,含有1 ,和不含有1;
含有1时,就是f[n-1],不含1时,就是f[n/2](拆分结果都除以2)的结果
综上结果,得出打表公式:
f[1]=1; f[2]=2;
f[n]=f[n-1] //n是奇数
f[n]=f[n-1]+f[n/2] //n是偶数
#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> using namespace std; int a[1000010]; int main() { int n; a[1] = 1; a[2] = a[3] = 2; a[4] = a[5] = 4; for(int i=6;i<=1000000;i++) { if(i%2==0) a[i] = (a[i-1] + a[i/2])%1000000000;//这里取i/2的值 else a[i] = a[i-1]%1000000000; } while(cin >> n) { cout << a[n] << endl; } return 0; }