Codeforces 225b well-known numbers

Http://codeforces.com/problemset/problem/225/ B

Idea: maintain a queue with a length of K and use the number kK-The number of bonacci instances is equal to the first K.K-Calculate the sum of bonacci numbers for all values not greater than S.K-Bonacci number, which can be recorded with a set, and then the number of S is closest to each greedy operation.

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 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         else43         {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_JUDGE57     freopen("in","r",stdin);58     #endif59     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 }