Numbers k-bonacci ( k is integer, k > 1) are a generalization of Fibonacci numbers and are determined as follows:
- F(k, n) = 0, for integer n, 1 ≤ n < k;
- F(k, k) = 1;
- F(k, n) = F(k, n - 1) + F(k, n - 2) + … + F(k, n - k), for integer n, n > k.
Note that we determine the k-bonacci numbers, F(k, n), only for integer values of n and k.
You’ve got a number s, represent it as a sum of several (at least two) distinct k-bonacci numbers.
Input
The first line contains two integers s and k (1 ≤ s, k ≤ 109; k > 1).
Output
In the first line print an integer m (m ≥ 2) that shows how many numbers are in the found representation. In the second line print m distinct integers a 1, a 2, …, a m. Each printed integer should be a k-bonacci number. The sum of printed integers must equal s.
It is guaranteed that the answer exists. If there are several possible answers, print any of them.
Examples
input
5 2
output
3
0 2 3
input
21 5
output
3
4 1 16
Solution:
#include <iostream> #include <iomanip> #include <stdio.h> #include <set> #include <vector> #include <map> #include <cmath> #include <algorithm> #include <memory.h> #include <string> #include <sstream> using namespace std; long long ff[111], f[111], ans[111]; int s, k, i, j, ka; void go(int v, int z) { if (z == 0) { if (ka == 1) ans[ka++] = 0; printf("%d\n",ka); for (i=0;i<ka-1;i++) printf("%d ",ans[i]); printf("%d\n",ans[ka-1]); exit(0); } if (v == 0) return; if (ff[v-1] >= z) go(v-1,z); if (f[v] <= z) { ans[ka++] = f[v]; go(v-1,z-f[v]); ka--; } } int main() { // freopen("in","r",stdin); // freopen("out","w",stdout); scanf("%d %d",&s,&k); f[0] = 1; for (i=1;i<=50;i++) { f[i] = 0; for (j=i-1;j>=i-k && j>=0;j--) f[i] += f[j]; if (f[i] > s) break; } ff[0] = 0; for (j=1;j<i;j++) ff[j] = ff[j-1]+f[j]; ka = 0; go(i-1,s); return 0; }