zoukankan      html  css  js  c++  java
  • K. Perpetuum Mobile

    The world famous scientist Innokentiy almost finished the creation of perpetuum mobile. Its main part is the energy generator which allows the other mobile's parts work. The generator consists of two long parallel plates with n lasers on one of them and n receivers on another. The generator is considered to be working if each laser emits the beam to some receiver such that exactly one beam is emitted to each receiver.

    It is obvious that some beams emitted by distinct lasers can intersect. If two beams intersect each other, one joule of energy is released per second because of the interaction of the beams. So the more beams intersect, the more energy is released. Innokentiy noticed that if the energy generator releases exactly k joules per second, the perpetuum mobile will work up to 10 times longer. The scientist can direct any laser at any receiver, but he hasn't thought of such a construction that will give exactly the required amount of energy yet. You should help the scientist to tune up the generator.

    Input

    The only line contains two integers n and k (1 ≤ n ≤ 200000, ) separated by space — the number of lasers in the energy generator and the power of the generator Innokentiy wants to reach.

    Output

    Output n integers separated by spaces. i-th number should be equal to the number of receiver which the i-th laser should be directed at. Both lasers and receivers are numbered from 1 to n. It is guaranteed that the solution exists. If there are several solutions, you can output any of them.

    Sample test(s)
    input
    4 5
    output
    4 2 3 1
    input
    5 7
    output
    4 2 5 3 1
    input
    6 0
    output
    1 2 3 4 5 6

    其实就是 求n的序列 的逆序对数为k;代码比较简单只是难以想到
    #include<iostream>
    using namespace std;
    int main(){
        int n;long long int k;
        while(cin>>n>>k){
               int flag=n,t=1;
            for(int i=1;i<=n;i++){
                if(k>=n-i){
                    k-=n-i;
                    cout<<flag--<<" ";
                }else{
                    cout<<t++<<" ";
                }
            }
            cout<<endl;
        }
    return 0;
    }
    View Code
  • 相关阅读:
    对于工程师责任和责任边界的认知
    windows消息机制与实例
    final阶段团队贡献分分配
    Final阶段用户调查报告
    对金州勇士团队的项目进行功能分解(加分作业)
    final review 报告
    final发布评论
    本周PSP
    final发布视频
    【转】mac版微信web开发者工具(小程序开发工具)无法显示二维码 解决方案
  • 原文地址:https://www.cnblogs.com/demodemo/p/4716150.html
Copyright © 2011-2022 走看看