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 mdistinct integers a1, a2, ..., am. 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.
题目背景是一个k—Fibonacci数列，也就是，第0，1项是1，然后后面的第i项为前面k项之和，即f[i]=f[i-1]+.....f[i-k],（i>=k+1），然后输入整数s，k，输出能使得加起来和为s的m（m>=2)个不同的k—Fibonacci数，1<=s<=10^9，k>1，考虑到k最小为2时，f[50]>=10^10，所以对于任意k，满足条件的整数不会超过10^9，只需要存储前50个就可以了。这样s依次减去小于它的最大Fibonacci值，直到s为0.
题目要求最少输出2个数，所以遇到恰好为Fibonacci数的s值，可以输出一个0。
代码：
1 #include<stdio.h> 2 #define max(a,b) ((a)>(b)?(a):(b)) 3 #define N 50 4 typedef long long ll; 5 ll f[N]; 6 int main(void) 7 { 8 int s,k; 9 int i,j,ct=0; 10 ll ans[N]; 11 scanf("%d%d",&s,&k); 12 f[0]=f[1]=1; 13 f[2]=2; 14 for(i=3;i<N;i++) 15 { 16 if(i>=k+1) 17 f[i]=2*f[i-1]-f[i-k-1];//i>=k+1,递推公式:f[i]=2*f[i-1]-f[i-k-1] 18 else for(j=i-1;j>=max(i-k,0);j--) 19 f[i]+=f[j];//否则f[i]为前面k项之和 20 } 21 for(i=N-1;i>0;i--) 22 { 23 if((s>=f[i])) 24 { 25 s-=f[i]; 26 ans[ct++]=f[i]; 27 } 28 if(s==0) 29 { 30 if(ct==1){ 31 printf("%d ",ct+1); 32 printf("0 "); 33 } 34 else printf("%d ",ct); 35 for(i=0;i<ct;i++) 36 printf("%I64d%c",ans[i],i==ct-1?' ':' '); 37 return 0; 38 } 39 } 40 return 0; 41 }