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  • Codeforces Round #256 (Div. 2) D. Multiplication Table(二进制搜索)

    转载请注明出处:http://blog.csdn.net/u012860063?viewmode=contents
    题目链接:http://codeforces.com/contest/448/problem/D

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    D. Multiplication Table
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    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 ann × 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 nm 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 test(s)
    input
    2 2 2
    
    output
    2
    
    input
    2 3 4
    
    output
    3
    
    input
    1 10 5
    
    output
    5
    
    Note

    2 × 3 multiplication table looks like this:

    1 2 3
    2 4 6

    题目意思是,从一个n*m的乘法表(不要问我乘法表是什么)中选出第k小数(同样的数字会计算多次)。

    比方例子 2 3 4

    乘法表为

    1 2 3

    2 3 4

    非减序列是:1, 2, 2, 3, 3, 4。第4个数字是3。所以输出3。

    代码例如以下:

    #include <iostream>
    #include <algorithm>
    using namespace std;
    typedef long long LL;
    LL n, m, k;
    LL min(LL a, LL b)
    {
    	return a<b?

    a:b; } LL check(LL x)//查找比x小的个数 { LL num = 0; for(int i = 1; i <= n; i++) { num+=min(m,x/i); } if(num >= k) return 1; else return 0; } int main() { while(cin>>n>>m>>k) { LL l = 0, r = n*m, ans = 0; while(l <= r) { LL mid = (l+r)>>1; if(check(mid)) { ans = mid; r = mid-1; } else l = mid+1; } cout<<ans<<endl; } return 0; }



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  • 原文地址:https://www.cnblogs.com/zfyouxi/p/5044150.html
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