数学--数论--HDU 5019 revenge of GCD】的更多相关文章

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--){…
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…
先放知识点: 莫比乌斯反演 卢卡斯定理求组合数 乘法逆元 快速幂取模 GCD of Sequence Alice is playing a game with Bob. Alice shows N integers a 1, a 2, -, a N, and M, K. She says each integers 1 ≤ a i ≤ M. And now Alice wants to ask for each d = 1 to M, how many different sequences b…
Describtion In mathematics, the greatest common divisor (gcd) of two or more integers, when at least one of them is not zero, is the largest positive integer that divides the numbers without a remainder. For example, the GCD of 8 and 12 is 4.-Wikiped…
Describtion First we define: (1) lcm(a,b), the least common multiple of two integers a and b, is the smallest positive integer that is divisible by both a and b. for example, lcm(2,3)=6 and lcm(4,6)=12. (2) gcd(a,b), the greatest common divisor of tw…
Problem Description Now given two kinds of coins A and B,which satisfy that GCD(A,B)=1.Here you can assume that there are enough coins for both kinds.Please calculate the maximal value that you cannot pay and the total number that you cannot pay. Inp…
This time I need you to calculate the f(n) . (3<=n<=1000000) f(n)= Gcd(3)+Gcd(4)+-+Gcd(i)+-+Gcd(n). Gcd(n)=gcd(C[n][1],C[n][2],--,C[n][n-1]) C[n][k] means the number of way to choose k things from n some things. gcd(a,b) means the greatest common di…
Your job is simple, for each task, you should output Fn module 109+7. Input The first line has only one integer T, indicates the number of tasks. Then, for the next T lines, each line consists of 6 integers, A , B, C, D, P, n. 1≤T≤200≤A,B,C,D≤1091≤P,…