Lucky Numbers

最新推荐文章于 2019-02-09 14:57:31 发布

原创最新推荐文章于 2019-02-09 14:57:31 发布 · 859 阅读

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本文详细阐述了幸运数字的生成算法，通过一系列步骤从给定整数序列中筛选出特定的数字序列，并最终计算出幸运数字的数量。利用C++编程实现算法，包括初始化过程、迭代删除操作以及输出结果等关键步骤。

Lucky Numbers

时间限制: 1 Sec 内存限制: 128 MB
提交: 22 解决: 9
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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

2030

样例输出

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.

来源

辽宁省赛2014

#include<iostream>
using namespace std;
int b[200];
void init(){
    int a[10005];
    for(int i=1;i<=10000;i++)
        a[i]=1;
    int s=0,sum=10000;
    int t=2,st=1; b[0]=2;
    while(sum>=t){
        for(int i=1;i<=10000;i++){
            if(a[i]>0){
                s++;
                if(s==t){
                    a[i]=0;
                    s=0;
                }
            }
        }
        s=0;
        sum=0;
        for(int i=1;i<=10000;i++)
            if(a[i]==1)
                sum++;
        for(int i=t+1;i<=10000;i++)
            if(a[i]==1){
                t=i;
                b[st++]=t;
                break;
            }
    }
}
int main(){
    init();     //把所有能取到的数保存在b[]里面
    int n;
    while(cin>>n){
        int i=0;
        while(n>=b[i]){
            n=n-n/b[i];
            i++;
        }
        cout<<n<<endl;
    }
    return 0;
}