CF_225B _Well-known Numbers

最新推荐文章于 2024-05-23 16:15:25 发布

rootial

最新推荐文章于 2024-05-23 16:15:25 发布

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本文链接：https://blog.youkuaiyun.com/rootial/article/details/38851045

本文深入探讨了k-Fibonacci数列的定义与性质，并提供了一个算法来表示给定数s为至少两个不同k-Fibonacci数的和。详细解释了如何通过迭代计算数列的值，以及如何从大到小逐步减去数列中的元素，直至s为0，确保输出的数列满足题目要求。

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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, ..., 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.

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\n",ct+1);
32                 printf("0 ");
33             }
34             else printf("%d\n",ct);
35             for(i=0;i<ct;i++)
36                 printf("%I64d%c",ans[i],i==ct-1?'\n':' ');
37             return 0;
38         }
39     }
40         return 0;
41 }