[atAGC045F]Division into Multiples】的更多相关文章

令$d=\gcd(a,b)$,可以发现$c|(ax+by)$等价于$lcm(c,d)|(ax+by)$,因此不妨令$c'=lcm(c,d)$,然后将$a$.$b$和$c$同时除以$d$ 接下来设$(a,c)=d_{1}$,根据整除的传递性有$d_{1}|(ax+by)$,由于$d_{1}|ax$,可得$d_{1}|by$,又因为$(b,d_{1})=1$,所以$d_{1}|y$ 因此,可以令$y'=\lfloor\frac{y}{d}\rfloor$,然后再将$a$和$c$同除以$d_{1}$,…
1.在python2 中导入未来的支持的语言特征中division(精确除法),即from __future__ import division ,当我们在程序中没有导入该特征时,"/"操作符执行的只能是整除,也就是取整数,只有当我们导入division(精确算法)以后,"/"执行的才是精确算法. 如: #python 2.7.6 Python 2.7.6 (default, Nov 10 2013, 19:24:18) [MSC v.1500 32 bit (In…
Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0. Example:Given a / b =…
Did you ever have the feeling that adding people doesn't help in software development? Did you ever think about the reason? And do you have any idea to make a change? Traditional software engineering emphasizes on division of work by modules, values…
#include<stdio.h> int main(void){ int n1, n2,n3; n1=333*(3+999)/2; n2=199*(5+995)/2; n3=66*(15+990)/2; printf("%d\n",n1+n2-n3); n1=getchar(); return 0; } Multiples of 3 and 5 Problem 1 If we list all the natural numbers below 10 that are m…
Contestants Division   Description In the new ACM-ICPC Regional Contest, a special monitoring and submitting system will be set up, and students will be able to compete at their own universities. However there’s one problem. Due to the high cost of t…
Multiples of 3 and 5 原题 题意如下: 找出N以内的3和5的倍数的和. 思路 1.刚看到觉得好弱智,直接遍历一遍不就OK了吗?但是第2和第3个测试用例报了TLE,超时. 2.然后想不出来了,搜了一下,发现有一个类似的题计算1000以内的3和5的倍数的和,用的是算出3的倍数的和加上5的倍数的和减去15的倍数的和,因为是直接计算所以不需要遍历,自然就快了.按照同样的思路,可以找到小于N的最大的3的倍数,5的倍数和15的倍数. 代码 int main(){ int t; cin >…
题目传送门 /* 暴力:对于每一个数都判断,是否数字全都使用过一遍 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <map> #include <set> #include <queue> usin…
GDC2016[全境封锁(Tom Clancy's The Division)]对为何对应Eye Tracked System,以及各种优点的演讲报告 原文 4Gamer編集部:松本隆一 http://www.4gamer.net/games/241/G024173/20160317137/         北美时间2016年3月16日,在旧金山举办的Games Developers Conference 2016上,进行了关于[全境封锁(Tom Clancy's The Division)]的…
Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0. Example: Given a / b =…