2014辽宁ACM省赛 Lucky Numbers

问题 G: Lucky Numbers

时间限制: 1 Sec 内存限制: 128 MB

提交: 17 解决: 8

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题目描述

A lucky number is made by the following rules:
Given a positive integer sequence {x | 1 <= x <= n}. From the first number, delete the last one in every 2 numbers. Select the minimum that has not been unused from the rest numbers. This number is xi. Then delete the last number in every xi numbers.
输入

There are several test cases, ended by the end of file.
Each test case has a positive integer n.(3<=n<=10000)
输出

Output the number of lucky numbers in one line.
样例输入
20 30
样例输出
6 8
提示

eg.1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

For the first time, delete the last number in every two numbers. The sequence become 1 3 5 7 9 11 13 15 17 19

For the second time, the minimum number that has never been used is 3. Delete the last number in every 3 numbers. The sequence become 1 3 7 9 13 15 19

For the third time, the minimum number that has never been used is 7. Delete the last number in every 7 numbers. The sequence becomes 1 3 7 9 13 15.

Then you cannot delete any numbers. There are 6 numbers left over. So the answer is 6.

一开始用容器模拟，结果超时了，于是后面发现数据范围其实也不大，然后就打表过了。。。。

这个是超时代码

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<set>
#include<vector>
using namespace std;
const int MAX=10001;
int vis[MAX];
int main()
{
freopen("out.txt","w",stdout);
for(int n=3;n<=10000;n++)
{
set<int> st;
vector<int> G;
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
{
st.insert(i);
G.push_back(i);
}
set<int>::iterator iter;
vector<int>::iterator itv;
int cnt=2;
int curp=0;
while(true)
{
int len=st.size();
int k=0;
for(int i=curp+cnt-1;i<len;i+=cnt)
{
// cout<<"a"<<i<<endl;
st.erase(G[i]);
vis[k++]=G[i];
}
for(int i=0;i<k;i++)
{
G.erase(lower_bound(G.begin(),G.end(),vis[i]));
}
//            for(iter=st.begin();iter!=st.end();iter++)
//            {
//                cout<<*iter<<" ";
//            }
//            cout<<endl;
bool flag=false;
for(iter=st.begin();iter!=st.end();iter++)
{
if(*iter>cnt&&*iter<=st.size())
{
cnt=*iter;
flag=true;
break;
}
}
if(flag==false)
{
break;
}
}
printf("%d,",st.size());
if(n%100==0) printf("\n");
}
return 0;
}

打表代码，大家都懂的，我就不贴了。。。。