又见GCD Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9984 Accepted Submission(s): 4157 Problem Description 有三个正整数a,b,c(0<a,b,c<10^6),当中c不等于b.若a和c的最大公约数为b,现已知a和b,求满足条件的最小的c. Input 第一行输入一个…
又见GCD Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 10031 Accepted Submission(s): 4185 Problem Description 有三个正整数a,b,c(0<a,b,c<10^6),当中c不等于b.若a和c的最大公约数为b,现已知a和b,求满足条件的最小的c. Input 第一行输入一…
又见GCD Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 18480 Accepted Submission(s): 7708 Problem Description 有三个正整数a,b,c(0<a,b,c<10^6),其中c不等于b.若a和c的最大公约数为b,现已知a和b,求满足条件的最小的c. Input 第一行输入一…
又见GCD Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9207 Accepted Submission(s): 3782 Problem Description 有三个正整数a,b,c(0<a,b,c<10^6),其中c不等于b.若a和c的最大公约数为b,现已知a和b,求满足条件的最小的c. Input 第一行输入一个…
这个题真的很水,但我竟然连错,在此警醒自己!!! 写代码改了东边,忘了西边,“认真”这两个字又被我吃了,打脸啪啪啪啪. #include<iostream> using namespace std; int gcd(int a, int b) { int t; while(b) { t = a; a = b; b = t%b; } return a; } int main() { int T; cin>>T; while(T--) { long long a, b; cin>…
Problem Description 有三个正整数a,b,c(0 import java.util.Scanner; public class Main{ public static void main(String[] args) { Scanner sc= new Scanner(System.in); int t =sc.nextInt(); while(t-->0){ int a =sc.nextInt(); int b =sc.nextInt(); int c=0; for(int…
#include<cstdio> int f1(int a,int b) //最大公约数 { ) return b; else return f1(b,a%b); } int f2(int a,int b) //最小公倍数 { int g; g=a*b/f1(a,b); return g; } int main() { int t,i; scanf("%d",&t); while(t--) { int a,b; scanf("%d%d",&…
第一道:Rightmost digit 求N^N次最后一个数字 快速幂mod10咯 代码如下: #include <cstdio> #define ll long long using namespace std; ; int qm(ll a,ll b) { ll res = ; while (b) { ) res = res * a % mod; a = a * a % mod; b >>= ; } return (int)res; } int main() { int T; s…
什么是GCD? Grand Central Dispatch或者GCD,是一套低层API,提供了一种新的方法来进行并发程序编写.从基本功能上讲,GCD有点像 NSOperationQueue,他们都允许程序将任务切分为多个单一任务然后提交至工作队列来并发地或者串行地执行.GCD比之 NSOpertionQueue更底层更高效,并且它不是Cocoa框架的一部分. 除了代码的平行执行能力,GCD还提供高度集成的事件控制系统.可以设置句柄来响应文件描述符.mach ports(Mach port 用于…
Grand Central Dispatch (GCD) Reference Grand Central Dispatch (GCD) comprises language features, runtime libraries, and system enhancements that provide systemic, comprehensive improvements to the support for concurrent code execution on multicore ha…
题目连接:http://www.spoj.com/problems/LGLOVE/ 题意:给出n个初始序列a[1],a[2],...,a[n],b[i]表示LCM(1,2,3,...,a[i]),即1~a[i]的最小公倍数 然后给出三种操作,注意:0<=i,j<n 0 i j p :a[i]~a[j]都加上p 1 i j :求LCM(b[i],b[i+1],...,b[j]) 2 i j :求GCD(b[i],b[i+1],...,b[j]) 思路: 求LCM(b[i],b[i+1],...,…
GCD Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7002 Accepted Submission(s): 2577 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y)…
