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:
    angularJS获取json数据(实战)
    HTML中使背景图片自适应浏览器大小
    实现table的单线边框的办法
    [转载]姑娘,你为什么要编程呢?
    可拖拽和带预览图的jQuery文件上传插件ssiuploader
    table布局的简单网页
    3D立体照片墙
    五一假期安排
    有点小迷惘
    一直都不明白,现在还是木有明白,那些人,那些事——残念
  • 原文地址:https://www.cnblogs.com/a972290869/p/4241968.html
Copyright © 2011-2022 走看看