http://codeforces.com/problemset/problem/225/B
思路：维护一个长度为k的队列，利用第k个k-bonacci数等于前面k个k-bonacci数的和，求出所有不大于s的k-bonacci数，可以用一个set来记录，然后每次贪最接近s的数。
View Code
1 /* 2 Author:Zhaofa Fang 3 Lang:C++ 4 */ 5 #include <cstdio> 6 #include <cstdlib> 7 #include <iostream> 8 #include <cmath> 9 #include <cstring> 10 #include <algorithm> 11 #include <string> 12 #include <vector> 13 #include <queue> 14 #include <stack> 15 #include <map> 16 #include <set> 17 using namespace std; 18 19 typedef long long ll; 20 const int INF = 2147483647; 21 22 set<int>su; 23 void cal(int s,int k) 24 { 25 queue<int>q; 26 q.push(1); 27 int sum=1; 28 su.insert(sum); 29 ll tmp; 30 while(1) 31 { 32 if(q.size() == k) 33 { 34 su.insert(sum); 35 tmp = (ll)(2*sum )- q.front(); 36 if(tmp > s)return; 37 su.insert(tmp); 38 q.push(sum); 39 sum = tmp; 40 q.pop(); 41 } 42 else 43 { 44 su.insert(sum); 45 tmp = (ll)sum * 2; 46 if(tmp > s)return ; 47 q.push(sum); 48 sum = tmp; 49 50 } 51 } 52 return; 53 } 54 int main() 55 { 56 #ifndef ONLINE_JUDGE 57 freopen("in","r",stdin); 58 #endif 59 int s,k; 60 while(~scanf("%d%d",&s,&k)) 61 { 62 cal(s,k); 63 int S[35]={0},top=1; 64 while(s) 65 { 66 set<int>::iterator it = su.lower_bound(s); 67 if(it != su.begin() && *it != s)it--; 68 S[top++]=*it; 69 s-=*it; 70 } 71 printf("%d\n",top); 72 for(int i=top-1;i>=0;i--)printf("%d ",S[i]); 73 puts(""); 74 } 75 return 0; 76 }