zoukankan      html  css  js  c++  java
  • (贪心) cf 435B

    B. Pasha Maximizes
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Pasha has a positive integer a without leading zeroes. Today he decided that the number is too small and he should make it larger. Unfortunately, the only operation Pasha can do is to swap two adjacent decimal digits of the integer.

    Help Pasha count the maximum number he can get if he has the time to make at most k swaps.

    Input

    The single line contains two integers a and k (1 ≤ a ≤ 1018; 0 ≤ k ≤ 100).

    Output

    Print the maximum number that Pasha can get if he makes at most k swaps.

    Sample test(s)
    input
    1990 1
    output
    9190
    input
    300 0
    output
    300
    input
    1034 2
    output
    3104
    input
    9090000078001234 6
    output
    9907000008001234
    B. Pasha Maximizes
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Pasha has a positive integer a without leading zeroes. Today he decided that the number is too small and he should make it larger. Unfortunately, the only operation Pasha can do is to swap two adjacent decimal digits of the integer.

    Help Pasha count the maximum number he can get if he has the time to make at most k swaps.

    Input

    The single line contains two integers a and k (1 ≤ a ≤ 1018; 0 ≤ k ≤ 100).

    Output

    Print the maximum number that Pasha can get if he makes at most k swaps.

    Sample test(s)
    input
    1990 1
    output
    9190
    input
    300 0
    output
    300
    input
    1034 2
    output
    3104
    input
    9090000078001234 6
    output
    9907000008001234

     其实 很简单,尽量交换前面的数字可以保证最优嘛

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<algorithm>
    #include<cstdlib>
    using namespace std;
    string s;
    int k;
    int main()
    {
         cin>>s>>k;
         for(int i=0;i<s.size();i++)
         {
               int b=i;
               for(int j=i+1;j<s.size()&&j-i<=k;j++)
               {
                     if(s[j]>s[b])
                            b=j;
               }
               k-=b-i;
               for(int j=b;j>i;j--)
                      swap(s[j],s[j-1]);
         }
         cout<<s<<endl;
         return 0;
    }
    

      

  • 相关阅读:
    go 学习成长之路
    多个ssh key 配置多个网址
    ubuntu 安装 gightingale
    关于open falcon 与nightingale 的一些调研
    kubeadm 命令简介
    windows kubectl 远程操作k8s
    ubuntu18.04 kuebadm 安装 k8s-1.15.9
    kubeadm 如何将节点加入集群
    centos7 kubeadm 搭建k8s
    docker 常用命令
  • 原文地址:https://www.cnblogs.com/a972290869/p/4241968.html
Copyright © 2011-2022 走看看