Let's consider a table consisting of n rows and n columns. The cell located at the intersection of i-th row and j-th column contains numberi × j. The rows and columns are numbered starting from 1. You are given a positive integer x. Your task is to c…

题目:click here 题意:看hint就懂了 分析:数论小题,在n0.5时间里求n的质因子数 #include <bits/stdc++.h> using namespace std; typedef long long ll; int n, x; ll ans; void solve() { ans = ; ; i*i<=x; i++ ) { if( i <= n ) { && x/i <= n ) { ; ; } } else break; } pr…

http://codeforces.com/contest/448/problem/D 题意:一个n×m的矩阵,a[i][j]=i*j; 然后把a数组排序,找出第k个数. 思路:1-n×m二分枚举,然后统计比小于等于x的数的个数与k相比较. #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #define maxn 1000100 #define ll…

传送门 冷静分析容易发现,我们只要能确定一个数的值,所有值也就可以确定了 确定一个数的值很容易,$a_ia_j=M_{i,j},a_ia_k=M_{i,k},a_ja_k=M_{j,k}$ 然后就可以得到 $a_i=\sqrt {M_{i,j}*M_{j,k}/M_{j,k}}$ ,然后这一题就做完了 #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #incl…

A. Multiplication Table Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/577/problem/A Description Let's consider a table consisting of n rows and n columns. The cell located at the intersection of i-th row and j-th column co…

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 -----------------------------------------------------------------------------------------------…

链接: https://codeforces.com/contest/1220/problem/B 题意: Sasha grew up and went to first grade. To celebrate this event her mother bought her a multiplication table M with n rows and n columns such that Mij=ai⋅aj where a1,-,an is some sequence of positi…

 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…

二分法判断答案 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 multiplicat…

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(…

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…

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…

http://acm.hdu.edu.cn/showproblem.php?pid=4951 2014多校 第八题 1008 2014 Multi-University Training Contest 8 Multiplication table Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 438    Accepted Submi…

题目链接:Codeforces 417E Square Table 题目大意:给出n和m.要求给出一个矩阵,要求每一列每一行的元素的平方总和是一个平方数. 解题思路:构造.依照 a a a b a a a b a a a b c c c d 的方式取构造,然后a,b,c,d的值用随机生成数去枚举,只是我认为用暴力也是能够的. #include <cstdio> #include <cstring> #include <cmath> #include <cstdli…

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…

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…

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 * nMultiplication Table, and a positive integer k, you need to retu…

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}…

题目地址:http://codeforces.com/contest/448/problem/D 当时是依照找规律做的,规律倒是找出来了,可是非常麻烦非常麻烦. . 看到前几名的红名爷们3分钟就过了,于是果断放弃了. .赛后才知道是用二分的方法做,知道了二分之后.剩下的就非常easy了..关键在于能不能想到用二分.. 代码例如以下: #include <iostream> #include <stdio.h> #include <string.h> #include &…

主题链接:http://codeforces.com/contest/448/problem/D 思路:用二分法 code: #include<cstdio> #include<cmath> #include<iostream> using namespace std; __int64 n,m,k; __int64 f(__int64 x) { __int64 res=0; for(__int64 i=1;i<=n;i++) { __int64 minn=min(…

C题, #include<cstdio> #include<cstring> #include<algorithm> #define maxn 5005 using namespace std; int num[maxn]; int rmq(int l,int r) { <<,tmp=l; for(int i=l;i<=r;i++) { if(ans>num[i]) { ans=num[i]; tmp=i; } } return tmp; } i…

题目链接 题意: n*m的一个乘法表,从小到大排序后,输出第k个数  (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m) 分析: 对于k之前的数,排名小于k:k之后的数大于,那么就能够採用二分. LL n, m, k; LL fun(LL goal) { LL t = 0, ret = 0; while (++t <= m) { ret += min(n, goal / t); } return ret; } LL bin(LL L, LL R, LL goal) { LL M, V…

二分!!! AC代码例如以下: #include<iostream> #include<cstring> #include<cstdio> #define ll long long using namespace std; ll n,m,k; ll work(ll a) { ll i,j; ll ans=0; for(i=1;i<=n;i++) { j=a/i; if(j>m) j=m; ans+=j; } return ans; } int main()…

题目链接:https://codeforces.com/problemset/problem/1099/E You are given an $n×m$ table, consisting of characters «A», «G», «C», «T». Let's call a table nice, if every $2×2$ square contains all four distinct characters. Your task is to find a nice table (…

题目链接:http://codeforces.com/problemset/problem/582/A 网上很多题解,就不说了,直接贴代码= = 官方题解: http://codeforces.com/blog/entry/20692 #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include &…

CF#256D:http://codeforces.com/contest/448/problem/D 题意:给你一个n*m的表格,第i行第j列的数据是i*j,然后询问第k小的数,这里的排序是不去重的. 题解:二分,每次判断当前的数是第几大,然后与k进行比较.这里统计这个数是第几大,要for一遍,一开始不知道怎么统计,看了别人的知道自己傻了.这里二分的时候不用判断这个数是否在这个表中,为什么不用,自己要好好体会. #include<iostream> #include<cstdio>…

http://www.codeforces.com/problemset/problem/22/B 题意:求出n*m的方格图中全是0的矩阵的最大周长 思路:枚举 #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> #include<iostream> ][],h[][]; ]; int read(){ ,f=;char ch=getchar(); ;ch=g…

听说这是一道$ Tourist$现场没出的题 Codeforces #662C 题意: 给定$n*m的 01$矩阵,可以任意反转一行/列($0$变$1$,$1$变$0$),求最少$ 1$的数量 $ n<=20 \ m<=100000$ $ Solution$ 考虑暴力 枚举每一行反转/不反转 预处理$ g(s)$表示某状态为$ s$的列的最少$ 1$的数量 显然$ g(s)=min(popcount(s),n-popcount(s))$ 枚举每行是否反转之后直接$ O(m)$计算即可 时间复杂…