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  • 【组队赛三】-E Binary Search cf448D

    Multiplication Table
    Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
    Submit

    Status

    Practice

    CodeForces 448D
    Description
    Bizon the Champion isn't just charming, he also is very smart.

    While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?

    Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.

    Input
    The single line contains integers n, m and k(1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).

    Output
    Print the k-th largest number in a n × m multiplication table.

    Sample Input
    Input
    2 2 2
    Output
    2
    Input
    2 3 4
    Output
    3
    Input
    1 10 5
    Output
    5
    Hint
    A 2 × 3 multiplication table looks like this:


    1 2 3
    2 4 6

    <span style="color:#3333ff;"><span style="color:#3333ff;background-color: rgb(255, 255, 255);">/*
    _______________________________________________________________________________________
    
           author    :   Grant yuan
           time      :   2014.7.21
           algorithm :   Binary Search
           explain   :   如果i*m<=aa,则会有m个数满足结果,否则会有aa/i个数满足结果 
    ________________________________________________________________________________________
    */
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    #include<queue>
    #include<functional>
    #define INF 999999999
    using namespace std;
    
    //long long a[100003][100003];
    long long k;
    long long l,r,mid;
    long long n,m;
    long long M;
    
    inline bool can(long long aa)
    {   long long sum=0;
        for(int i=1;i<=n;i++)
             {
                 if(i*m<=aa)
                    sum+=m;
                 else sum+=aa/i;
             }
       if(sum>=k)
           return true;
       return false;
    }
    
    int main()
    {   M=0;
        cin>>n>>m>>k;
            long long ans=1;
            l=1;r=n*m;
            while(l<=r){
                mid=(long long)((l+r)*0.5);
                if(can(mid))
                {  ans=mid;
                    r=mid-1;
                }
                else
                    l=mid+1;
            }
              cout<<ans<<endl;
    }
    </span></span>



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  • 原文地址:https://www.cnblogs.com/codeyuan/p/4254490.html
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