Copy Code Copy Command Solve the Diophantine equation, 30x+56y=8 for x and y. Find the greatest common divisor and a pair of Bézout coefficients for 30 and 56. Get [g,u,v] = gcd(30,56) g = 2 u = -13 v = 7 u and v satisfy the Bézout's identity, (30*u) + (...
DispatchQueue.main.async { // Run async code on the main queue}但是如果你再深入研究一下,会发现GCD还有一套你并不知道的并且非常强大的API和功能.这周,让我们将目光越过async {},进一步去发现它如何为其他大量的Foundation API提供更简单的(更灵活更具Swifty特色的)方法.使用...
然后贪心,要使得递增序列的公因子最大,先从大到小枚举因数 n/i ,没找到满足的因数再从小到大枚举因数 i。若当前枚举的gcd为x,若1x+2x+3x+...+kx<=n,那么必能把n分成k个递增的x的倍数。 #include <iostream>#include<cstdio>#include<cmath>usingnamespacestd;longlongn,k;boolcheck(longlongx) {//...
// 延迟2秒执行: doubledelayInSeconds = 2.0; dispatch_time_t popTime = dispatch_time(DISPATCH_TIME_NOW, delayInSeconds * NSEC_PER_SEC); dispatch_after(popTime, dispatch_get_main_queue(), ^(void){ // code to be executed on the main queue after delay }); 3、自定义GCD 可以使用dispatc...
codeforce 803C Maximal GCD(构造) 题意:对于给定n,k 求一个有k个数的上升序列,序列的和为n。且序列的最大公约数最大。 首先1s的时间 n k的数量级都是1e10 ,因此肯定要降低复杂度。 假设存在一个这样的序列,设序列的最大公约数为gcd,那么这个序列一定可以为 1∗gcd,2∗gcd,3∗gcd...(k−...
There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of ...Codeforces 892A Greed 求一个sum并与最大的两个容量之和作比较即可。 数据范围爆int,改用ll。...codeforces Pride 892C 题目链接:...
only execute the code once even in the presence of multithreaded calls. - parameter token: A unique reverse DNS style name such as com.vectorform.<name> or a GUID - parameter block: Block to execute once */publicclassfunconce(token:String,block:@noescape(Void)->Void) { ...
But before we go through all the specific topics of this post, let’s talk a bit about some specific concepts. First off, the dominating phrase in GCD is thedispatch queue. A queue is actually a block of code that can be executedsynchronouslyorasynchronously, either on the main or on a...
To check whether something real can be implemented in 'pure' Swift. Meaning, without using any Objective-C Cocoa classes (no NS'ism). Or in other words: Can you use Swift without writing all the 'real' code in wrapped Objective-C? :-)...
// code to be executed once }); // 延迟2秒执行: double delayInSeconds = 2.0; dispatch_time_t popTime = dispatch_time(DISPATCH_TIME_NOW, delayInSeconds * NSEC_PER_SEC); dispatch_after(popTime, dispatch_get_main_queue(), ^(void){ ...