https://www.hackerrank.com/contests/w8/challenges/john-and-gcd-list 简单题,GCD和LCM. #include <vector> #include <iostream> using namespace std; int gcd(int a, int b) { if (a % b == 0) { return b; } else { return gcd(b, a % b); } } int lcm(int a, i…

题解:tags:[预处理][gcd]故事背景:光头钻进了茫茫人海.这是一个典型の通过前缀后缀和来降低复杂度的问题.先用pre数组与suf数组分别维护前缀gcd和后缀gcd.如果 a[i] % gcd(pre[i-1], suf[i+1]) != 0;那么光头就从人群中钻出来了!gcd(pre[i-1],suf[i+1])就是我们要的答案.注意n=1时的特判! code: #include <iostream> #include <algorithm> #include <ve…

题意 题目链接 Sol 一道咕咕咕了好长时间的题 题解可以看这里 #include<bits/stdc++.h> #define LL long long using namespace std; const int MAXN = 1e7 + 5e6 + 10, mod = 1e9 + 7, mod2 = 1e9 + 6; int N, M, vis[MAXN], prime[MAXN], mu[MAXN], f[MAXN], tot; int add(int x, int y) { if(x…

原题地址 简单动归+素数判定,没用筛法也能过 代码: #include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> #include <cstring> using namespace std; #define MAX_N 64 #define MAX_M 1000000 int T, N; int cnt[M…

https://www.hackerrank.com/contests/w7/challenges/die-hard-3 首先,发现c <= max(a, b) 而且 c = aX + bY时有解.然后根据扩展欧几里得算法知道c % gcd(a, b) == 0时有解. #include <vector> #include <iostream> using namespace std; int gcd(int a, int b) { if (a % b == 0) { ret…

Grand Central Dispatch (GCD)是Apple开发的一个多核编程的较新的解决方法.它主要用于优化应用程序以支持多核处理器以及其他对称多处理系统.它是一个在线程池模式的基础上执行的并行任务.在Mac OS X 10.6雪豹中首次推出,也可在IOS 4及以上版本使用. 设计 GCD是一个替代诸如NSThread等技术的很高效和强大的技术.GCD完全可以处理诸如数据锁定和资源泄漏等复杂的异步编程问题.GCD的工作原理是让一个程序,根据可用的处理资源,安排他们在任何可用的处理器核心…

1 import fractions, functools, sys if __name__ == '__main__': T = int(sys.stdin.readline()) for _ in range(T): N = int(sys.stdin.readline()) A = list(map(int, sys.stdin.readline().split())) gcd = functools.reduce(fractions.gcd, A) # reduce函数的理解 print…

Play with GCD 题目连接: https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/play-with-gcd Description Minka is very smart kid who recently started learning computer programming. He learned how to calculate the Greatest Common Divisor (GC…

https://www.hackerrank.com/contests/infinitum16-firsttimer/challenges/solve-equations 给定一条方程a*x + b*y = c 保证有解的情况下,我们要求一个点,满足x > 0且 这个点到原点的欧几里德距离最短 根据扩展欧几里德算法,我们能求得一组(x1,y1)满足x1 > 0的解,第一个x>0的解. 然后通解就是x2 = x1 + b/abgcd*k  y2 = y1 - a/abgcd*k 如果(x2…

Red John has committed another murder. But this time, he doesn't leave a red smiley behind. What he leaves behind is a puzzle for Patrick Jane to solve. He also texts Teresa Lisbon that if Patrick is successful, he will turn himself in. The puzzle be…