任意门:http://acm.hdu.edu.cn/showproblem.php?pid=2588 GCD Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3608    Accepted Submission(s): 1954 Problem Description The greatest common divisor GCD(a,…

GCD Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u Submit Status Description The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For examp…

题目大意:给定N,M, 求1<=X<=N 且gcd(X,N)>=M的个数. 题解:首先,我们求出数字N的约数,保存在约数表中,然后,对于大于等于M的约数p[i],求出Euler(n/p[i]),累计就是答案.因为对于每一个大于等于m的约数,GCD(N,t*p[i])=p[i]>=m(t与p[i]互质),所以n除以p[i]的欧拉函数的和就是答案. #include <cstdio> int T,cnt,p[10000],n,m,i; int Eular(int n){ i…

GCD Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3559    Accepted Submission(s): 1921 Problem Description The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes writ…

GCD(一) 题目: The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6. (a,b) can be easily found by the Euclidean algorithm. Now Carp is conside…

Description The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6. (a,b) can be easily found by the Euclidean algorithm. Now Carp is consid…

题目大意: 求1~N中与N的最大公约数大于M的个数 思路: 这个题是不是可以想到暴力枚举??对于每一组数据枚举与他的最大公约数大于m的数的个数. 是,这种做法没错误,但是保准你T成狗.... 我们至少要找一个不T的做法吧...我们考虑gcd这样一个性质gcd(x,y)=m则gcd(x/m,y/m)=1;我们就可以轻易的发现在这个地方的x/m不就是我们要求的第一个式子中的x吗??这样我们就只需要统计这样的x/m的个数不就好了吗?! 这样显然就可以知道,这不就是欧拉函数吗?! 是的,那我们就来尝试一…

GCD Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2742    Accepted Submission(s): 980 Problem Description Give you a sequence of N(N≤100,000) integers : a1,...,an(0<ai≤1000,000,000). There ar…

求满足$1<=X<=N ,(X,N)>=M$的个数,其中$N, M (2<=N<=1000000000, 1<=M<=N)$. 首先,假定$(x, n)=m$,那么 $(\frac{x}{n},\frac{n}{m})=1$,故$$ans=\sum_{i=m}^{n}\varphi(\frac{n}{i})$$ ん?遅い! $$\sum_{i=m}^{n}\varphi(\frac{n}{i})=\sum\limits_{d|n}{\varphi(\frac{n}…

输入 N 和 M (2<=N<=1000000000, 1<=M<=N), 找出所有满足1<=X<=N 且 gcd(X,N)>=M 的 X 的数量. Input第一行输入样例数T (T <= 100)每个样例输入两个整数N , M. (2<=N<=1000000000, 1<=M<=N)Output对于每组样例,输出一个整数,表示满足条件的X的数量.Sample Input 3 1 1 10 2 10000 72 Sample Out…