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  • (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;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/a972290869/p/4241968.html
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