Description 给nn个小于pp的非负整数a1,-,na1,-,n,问有多少对(i,j)(1≤i<j≤n)(i,j)(1≤i<j≤n)模pp在意义下满足1ai+aj≡1ai+1aj1ai+aj≡1ai+1aj,即这两个数的和的逆元等于这两个数的逆元的和,注意0没有逆元 Input 第一行一整数TT表示用例组数,每组用例首先输入一整数nn表示序列长度和一素数pp表示模数,之后输入nn个非负整数a1,-,n(1≤T≤5,1≤n≤2×105,2≤p≤1018,0≤a1,-,n<p)a1…
题目链接 Problem Description There are n nonnegative integers a1-n which are less than p. HazelFan wants to know how many pairs i,j(1≤i<j≤n) are there, satisfying 1ai+aj≡1ai+1aj when we calculate module p, which means the inverse element of their sum equ…
http://acm.hdu.edu.cn/showproblem.php?pid=6128 题意:有一个a数列,并且每个数都小于p,现在要求有多少对$(i,j)$满足$\frac{1}{a_i+a_j} \equiv \frac{1}{a_i}+\frac{1}{a_j} \mod p$,0没有逆元. 思路:根据同余的性质对式子进行化简,在一个同余式两边同时做加法.减法或乘法仍保持同余. 先可以化简为 $1 \equiv 1+\frac{a_j}{a_i}+1+\frac{a_i}{a_j}…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6128 题意:给你n个数,问你有多少对i,j,满足i<j,并且1/(ai+aj)=1/ai+1/aj 在%p意义下. 解法:官方题解说是用二次剩余来解,但是我并不会这玩意了.在网上看到一位大佬没有二次剩余直接通过推公式做出了这题,真是神奇.http://www.cnblogs.com/bin-gege/p/7367337.html  将式子通分化简后可得(ai2+aj2+ai*aj)%p=0 .然后两…
Giving the N, can you tell me the answer of F(N)? Input Each test case contains a single integer N(1<=N<=10^9). The input is terminated by a set starting with N = 0. This set should not be processed. Output For each test case, output on a line the v…
A cubic number is the result of using a whole number in a multiplication three times. For example, 3×3×3=27 so 27 is a cubic number. The first few cubic numbers are 1,8,27,64 and 125. Given an prime number p. Check that if p is a difference of two cu…
Diophantus of Alexandria was an egypt mathematician living in Alexandria. He was one of the first mathematicians to study equations where variables were restricted to integral values. In honor of him, these equations are commonly called diophantine e…
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,…
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…
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…