/************************************************************************* > File Name: L.cpp > Author: LyuCheng > Created Time: 2017-12-03 17:31 > Description: 题意:首先有一个乘法表,然后给你一个n*m的矩阵,里面有不确定的数,然 后问你,这个表可不可能是乘法表的一部分 思路:暴力判断每个点 ***************…

给定一副表,问其是否合法. 思路:当全部是?的时候,是合法的. 如果不是,那么,就找到一个数字,把它拆成若干个a*b的形式,去判断其它点是否合法即可. 拆分数字的时候,只需要枚举到sqrt(n),因为肯定是两个小于sqrt n的数相乘得到的结果. 比如6=1*6 6=2*3 注意分解后,考虑调换顺序是否合法即可. #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath>…

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 tabl…

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

 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 t…

题目链接:D. Multiplication Table 题意: 给出N×M的乘法矩阵要你求在这个惩罚矩阵中第k个小的元素(1 ≤ n, m ≤ 5·10^5; 1 ≤ k ≤ n·m). 题解: n和m最大都是5e5那矩阵最大就有2e11不能够暴力,其实这里就应该想到要用二分做的,但是我做题的时候脑抽@.@想要推规律,然后就决决了.那么讲一下二分怎么做,就先简单的二分答案在(1-n*m)中二分,然后cheak函数中找所有矩阵中比所给值小的元素,个数大于等于k为成立条件.复杂度(O(n*log(…

Nearly every one have used the Multiplication Table. But could you find out the k-th smallest number quickly from the multiplication table? Given the height m and the length n of a m * n Multiplication Table, and a positive integer k, you need to ret…

题目:https://leetcode.com/problems/kth-smallest-number-in-multiplication-table/description/ 668. Kth Smallest Number in Multiplication Table Nearly every one have used the Multiplication Table. But could you find out the k-th smallest number quickly fr…

目录 题目 思路 \(Code\) 题目 CF448D Multiplication Table 思路 二分答案.这个矩阵的每一排都是递增的,所以二分\(ans\),去计算有多少个数等于\(ans\),有多少个数小于\(ans\),如果小于\(ans\)的数不多于\(k-1\)个并且小于等于\(ans\)的数不少于\(k\)个,那么当前\(ans\)就是答案. Q:如何计算小于\(ans\)的数的个数? A: \[\sum_{i=1}^{n}min(\lfloor \frac{ans-1}{i}…

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 i…