One Chance To Be The God Of Chrysanthemum

One Chance To Be The God Of Chrysanthemum

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 59   Accepted Submission(s) : 3

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Problem Description

　　Last week xysDavid had gone for an interview. The interviewer asked him a funny question. Here is it.
　　Given a number n, your mission is to find a set, that every number between 1 and n (inclusive) can be the sum of some elements in this set.
　　xysDavid seckilled it soon after reading the statement. If you can solve the problem too, xysDavid will accept you to be his apprentice, who can get the honorary title 'the god of chrysanthemum'.
　　Can you finish it?

Input

　　There is a number T (1<=T<=50) in the first line, indicating the number of test cases.
　　There is a number n in each line. (1<=n<=10^18);

Output

　　You should output "Case #D:" first, which "D" means the index of the test case.
　　Then output all the numbers in the set in one line in increasing order. Use one space to separate two numbers. Please note that no more spaces in the end of a line.
　　If there are multiple answers, output the one with smallest number of elements. If it is still a tie, output the one with the smallest lexicographical order.
　　The comparison between two sequences uses lexicographical ordering: first, the first two numbers are compared, and if they differ this determines the outcome of the comparison; if they are equal, the next two numbers are compared, and so on, until either sequence is exhausted. If all numbers of two sequences compare equal, the sequences are considered equal. If one sequence is an initial subsequence of the other, the shorted sequence is considered the smaller one.

Sample Input

213

Sample Output

Case #1:1Case #2:1 2

Source

华南师范大学2012年ACM程序设计竞赛（初赛+决赛无尽版）

 

二分的思想

题意是：得到一个集合，使1到n都能用该集合内的某些元素和来表示。如果求得的集合有多个，输出元素个数最少的一个，如果还是有，输出字典序最小的集合。

 

#include<iostream>#include<cstdio>#include<set>using namespace std;#define LL __int64set<LL> q;set<LL>::iterator it;LL solve(LL n) {if(n==1){q.insert(1);return 1;}else{LL a=solve(n>>1);//a以下的数都能表示 LL b=n-a;//剩下的数b while(q.count(b))//找到不重复的最小值 {b++;}q.insert(b);return b+a;//返回a+b以下都能表示 }}int main(){int t;int cnt=1;cin>>t;while(t--){LL n;cin>>n;q.clear();printf("Case #%d:\n",cnt++);solve(n);it=q.begin();printf("%I64d",*it);for(it++;it!=q.end();it++)printf(" %I64d",*it);cout<<endl;}return 0;}