题目链接: xiaoxin juju needs help Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 491 Accepted Submission(s): 142 Problem Description As we all known, xiaoxin is a brilliant coder. He knew **pal…
GCD is Funny 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5902 Description Alex has invented a new game for fun. There are n integers at a board and he performs the following moves repeatedly: He chooses three numbers a, b and c written at the boa…
GCD Expectation Time Limit: 4 Seconds Memory Limit: 262144 KB Edward has a set of n integers {a1, a2,...,an}. He randomly picks a nonempty subset {x1, x2,…,xm} (each nonempty subset has…
GCD and LCM Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4497 Description Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) = G and…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5019 Problem Description In mathematics, the greatest common divisor (gcd), also known as the greatest common factor (gcf), highest common factor (hcf), or greatest common measure (gcm), of two or more i…
Cake Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2077 Accepted Submission(s): 1078 Problem Description 一次生日Party可能有p人或者q人参加,现准备有一个大蛋糕.问最少要将蛋糕切成多少块(每块大小不一定相等),才能使p人或者q人出席的任何一种情况,都能平均将蛋糕分食…
Luogu3455:莫比乌斯反演进行GCD计数 莫比乌斯反演就是用来解决这一类问题的,通常f函数是要求的那个,F函数是显然的 这样利用F的结果就可以推出来f的结果 在计算结果的时候整除分快儿一下就可以很快了 #include<cstdio> #include<algorithm> using std::min; ; int cnt; long long ans; bool vis[maxn]; int mu[maxn],sum[maxn]; long long prim[maxn]…
GCD & LCM Inverse Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9928 Accepted: 1843 Description Given two positive integers a and b, we can easily calculate the greatest common divisor (GCD) and the least common multiple (LCM) of a a…
洛谷P2398 GCD SUM 题目描述 for i=1 to n for j=1 to n sum+=gcd(i,j) 给出n求sum. gcd(x,y)表示x,y的最大公约数. 输入输出格式 输入格式: n 输出格式: sum 输入输出样例 输入样例#1: 2 输出样例#1: 5 说明 数据范围 30% n<=3000 60% 7000<=n<=7100 100% n<=100000 Solution 这道题的做法貌似很多...如果你同时会狄利克雷卷积和莫比乌斯反演的话也可以强…
求最大公约数利用辗转相除法: long long gcd(long long a,long long b) { ) return a; else return gcd(b,a%b); } 求最小公倍数时,利用两数的乘积除以这两个数的最大公约数即可: long long lcm(long long a,long long b) { long long tmp=a*b; tmp=tmp/gcd(a,b); return tmp; } 完整代码如下,输入两个数n和m,输出这两个数的最大公约数和最小公倍…