题目链接 Problem Description Multiple query, for each n, you need to get $$$$$$ \sum_{i=1}^{n} \sum_{j=1}^{i-1}{ [gcd(i + j, i - j) = 1]} $$$$$$ Input On the first line, there is a positive integer T, which describe the number of queries. Next there are…

题意 给一个\(n\),计算 \[\sum_{i=1}^{n}\sum_{j=1}^{i-1}[gcd(i + j, i - j) = 1]\] 题解 令\(a = i - j\) 要求 \[\sum_{i=1}^{n}\sum_{j=1}^{i-1}[gcd(i + j, i - j) = 1]\] 即求 \[\sum_{i=1}^{n}\sum_{a=1}^{i-1}[gcd(2*i - a, a) = 1]\] 根据\(gcd\)的性质,即 \[\sum_{i=1}^{n}\sum_{a=…

1.HDU 2824   The Euler function 2.链接:http://acm.hdu.edu.cn/showproblem.php?pid=2824 3.总结:欧拉函数 题意:求(a,b)间的欧拉函数值的和. #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio>…

GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4141    Accepted Submission(s): 1441 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y)…

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…

输入a b c d k求有多少对x y 使得x在a-b区间 y在c-d区间 gcd(x, y) = k 此外a和c一定是1 由于gcd(x, y) == k 将b和d都除以k 题目转化为1到b/k 和1到d/k 2个区间 如果第一个区间小于第二个区间 讲第二个区间分成2部分来做1-b/k 和 b/k+1-d/k 第一部分对于每一个数i 和他互质的数就是这个数的欧拉函数值 全部数的欧拉函数的和就是答案 第二部分能够用全部数减去不互质的数 对于一个数i 分解因子和他不互质的数包括他的若干个因子 这个…

GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9046    Accepted Submission(s): 3351 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y…

GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5064    Accepted Submission(s): 1818 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y)…

欧拉函数 欧拉函数是指:对于一个正整数n,小于n且和n互质的正整数(包括1)的个数,记作φ(n) . 通式:φ(x)=x*(1-1/p1)(1-1/p2)(1-1/p3)*(1-1/p4)-..(1-1/pn),其中p1, p2--pn为x的所有质因数,x是不为0的整数.φ(1)=1(唯一和1互质的数就是1本身). 对于质数p,φ(p) = p - 1.注意φ(1)=1. 欧拉定理:对于互质的正整数a和n,有aφ(n) ≡ 1 mod n. 欧拉函数是积性函数--若m,n互质,φ(mn)=φ(m…

GuGuFishtion \[ Time Limit: 1500 ms\quad Memory Limit: 65536 kB \] 题意 给出定义\(Gu(a, b) = \frac{\phi(ab)}{\phi(a)\phi(b)}\) 求出\(\sum_{a=1}^{m}\sum_{b=1}^{n}Gu(a,b) (mod p)\) 思路 首先对于欧拉函数,我们知道欧拉函数的朴素式子为:\(\phi(n) = n*(1-\frac{1}{p1})*(1-\frac{1}{p2}) * ..…