GCD 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=1695 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) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be…

题目链接 求 $ x\in[1, a] , y \in [1, b] $ 内 \(gcd(x, y) = k\)的(x, y)的对数. 问题等价于$ x\in[1, a/k] , y \in [1, b/k] $ 内 \(gcd(x, y) = 1\) 的(x, y)的对数. 假设a < b, 那么[1, a/k]这部分可以用欧拉函数算. 设 \(i\in (a/k, b/k]\), (a/k, b/k]这部分可以用容斥算, 用a/k减去[1, a/k]里面和i不互质的数的个数. 具体看代码.…

题意:求区间1<=i<=b与区间1<=j<=d之间满足gcd(i,j) = k 的数对 (i,j) 个数.(i,j)与(j,i) 算一个. 分析:gcd(i,j)=k可以转化为gcd(i/k,j/k)=1.枚举每个1<=i<=b/k 的 i,用容斥原理统计区间[1,d]中与其互素的个数.需要预处理筛出2~1e5中每个数的质因子. *注意当k=0时,数对不存在. #include<bits/stdc++.h> using namespace std; type…

题意:给定区间和n,求区间中与n互素的数的个数, . 思路:利用容斥定理求得先求得区间与n互素的数的个数,设表示区间中与n互素的数的个数, 那么区间中与n互素的数的个数等于.详细分析见求指定区间内与n互素的数的个数 容斥原理 AC代码 #include <cstdio> #include <cmath> #include <cctype> #include <bitset> #include <algorithm> #include <cs…

Frogs Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5514 Description There are m stones lying on a circle, and n frogs are jumping over them.The stones are numbered from 0 to m−1 and the frogs are numbered fro…

GCD Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 3   Accepted Submission(s) : 2 Problem Description The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b…

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: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7529    Accepted Submission(s): 2773 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…

http://acm.hdu.edu.cn/showproblem.php?pid=6053 题意:给定一个数组,我们定义一个新的数组b满足bi<ai 求满足gcd(b1,b2....bn)>=2的数组b的个数 题解:利用容斥定理.我们先定义一个集合f(x)表示gcd(b1,b2...bn)为x倍数的个数(x为质数),我们在定义一个数mi为数组中的最小值,那么集合{f(2)Uf(3)....f(n)}就是我们想要的答案.f(x)=(a1/x)*(a2/x)*.....(ai/x),直接累加肯定…

How many integers can you find Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3315    Accepted Submission(s): 937 Problem Description   Now you get a number N, and a M-integers set, you shoul…