RXD is a good mathematician. One day he wants to calculate: output the answer module 109+7. p1,p2,p3-pk are different prime numbers Input There are several test cases, please keep reading until EOF. There are exact 10000 cases. For each test case, th…

题目链接 Problem Description RXD is a good mathematician. One day he wants to calculate: ∑i=1nkμ2(i)×⌊nki−−−√⌋ output the answer module 109+7. 1≤n,k≤1018 μ(n)=1(n=1) μ(n)=(−1)k(n=p1p2-pk) μ(n)=0(otherwise) p1,p2,p3-pk are different prime numbers Input Th…

比赛时候面向过题队伍数目 打表- - 看了题解发现确实是这么回事,分析能力太差.. /* HDU 6063 - RXD and math [ 数学,规律 ] | 2017 Multi-University Training Contest 3 题意: 求 Σ μ(i)^2 * sqrt( n^k/i ) [ 1 <= i<= n^k ] n,k <= 1e18 分析: 首先 μ(i) 为莫比乌斯函数,若 i 是完全平方数的倍数则 μ(i) = 0 ,否则 μ(i) = ±1 所以只有不是…

RXD and math Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 568    Accepted Submission(s): 306 Problem Description RXD is a good mathematician.One day he wants to calculate: ∑i=1nkμ2(i)×⌊nki−…

Bryce1010模板 http://acm.hdu.edu.cn/showproblem.php?pid=6063 打表发现规律是n^k #include <iostream> #include<string.h> #include<cmath> using namespace std; #define ll long long const int MOD=1e9+7; const int MAXN=1e6; bool check[MAXN+10]; int prim…

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,…

Problem Description When Teddy was a child , he was always thinking about some simple math problems ,such as "What it's 1 cup of water plus 1 pile of dough .." , "100 yuan buy 100 pig" .etc.. One day Teddy met a old man in his dream ,…

Ignatius's puzzle Problem Description Ignatius is poor at math,he falls across a puzzle problem,so he has no choice but to appeal to Eddy. this problem describes that:f(x)=5x13+13*x5+ka*x,input a nonegative integer k(k<10000),to find the minimal none…

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…

传送门 •题意 一直整数$a,b$,有 $\left\{\begin{matrix}x+y=a\\ LCM(x*y)=b \end{matrix}\right.$ 求$x,y$ •思路 解题重点:若$gcd(p,q)=1$,则$gcd(p+q,pq)=1$ 设$gcd(x,y)=g$,令$p=\frac{x}{g},q=\frac{y}{g}$,$p,q$互素 则$\left\{\begin{matrix}x+y=p*g+q*g=(p+q)g=a\\ LCM(x,y)=\frac{xy}{g}=…