题意: 给你两个数:X和Y  .输出它们的第K大公约数.若不存在输出 -1 数据范围: 1 <= X, Y, K <= 1 000 000 000 000 思路: 它俩的公约数一定是gcd(X,Y)的因数.(把它俩分解成质因数相乘的形式就可以看出) 故找出gcd(x,y)所有的因数,从大到小排序,输出第K个即可. 代码: #include <cstdio> #include <iostream> #include <string.h> #include &l…

Revenge of GCD 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 integers (when at least one of them is…

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…

题解:筛出约数,然后计算即可. #include <cstdio> #include <algorithm> typedef long long LL; LL a1[1000005],a2[1000005],x,y,k,g; int cnt1,cnt2,T; LL gcd(LL a,LL b){if(b==0)return a;else return gcd(b,a%b);} int main(){ scanf("%d",&T); while(T--){…

大水题 #include<time.h> #include <cstdio> #include <iostream> #include<algorithm> #include<math.h> #include <string.h> #include<vector> #include<queue> typedef long long int ll; using namespace std; ll f[]; int…

Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 221    Accepted Submission(s): 58 Problem Description This is a simple problem. The teacher gives Bob a list of probl…

CA Loves GCD  Accepts: 64  Submissions: 535  Time Limit: 6000/3000 MS (Java/Others)  Memory Limit: 262144/262144 K (Java/Others) 问题描述 CA喜欢是一个热爱党和人民的优秀同♂志,所以他也非常喜欢GCD(请在输入法中输入GCD得到CA喜欢GCD的原因). 现在他有N个不同的数,每次他会从中选出若干个(至少一个数),求出所有数的GCD然后放回去. 为了使自己不会无聊,CA…

Revenge of GCD 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 integers (when at least one of them is not zero), is the l…

题目链接:hdu 4983 Goffi and GCD 题目大意:求有多少对元组满足题目中的公式. 解题思路: n = 1或者k=2时:答案为1 k > 2时:答案为0(n≠1) k = 1时:须要计算,枚举n的因子.令因子k=gcd(n−a,n, 那么还有一边的gcd(n−b,n)=nk才干满足相乘等n.满足k=gcd(n−a,n)的a的个数即为ϕ(n/s),欧拉有o(n‾‾√的算法 #include <cstdio> #include <cstring> #include…

HDU 4983 Goffi and GCD 思路:数论题.假设k为2和n为1.那么仅仅可能1种.其它的k > 2就是0种,那么事实上仅仅要考虑k = 1的情况了.k = 1的时候,枚举n的因子,然后等于求该因子满足的个数,那么gcd(x, n) = 该因子的个数为phi(n / 该因子),然后再利用乘法原理计算就可以 代码: #include <cstdio> #include <cstring> #include <cmath> typedef long lo…