题目链接:https://vjudge.net/problem/UVA-11426 题目大意: 给出整数n∈[2,4000000],求解∑gcd(i,j),其中(i,j)满足1≤i<j≤n. 的确没有想到是欧拉函数,这怎么会想到欧拉函数呢?  又不是要我们求所有gcd为1的个数  那些gcd不为1的怎么办呢?  当时怎么就没想到呢  除过去不就变为1了吗  自己是真的菜... 还是要多做题,把思维开阔起来!!! 思路在代码中  直接看代码: /** 欧拉函数三个性质 是素数的话 欧拉函数值等于它…

题目 传送门:QWQ 分析 仪仗队 呃,看到题后感觉很像上面的仪仗队. 仪仗队求的是$ gcd(a,b)=1 $ 本题求的是$ gcd(a,b)=m $ 其中m是质数 把 $ gcd(a,b)=1 $ 变形成 $ gcd(a,b)*m=m $ 然后在n的范围内枚举一下,使 $ gcd(a,b)*m <= n $ 再像仪仗队那样用欧拉函数搞一搞. 没了. 代码 #include <bits/stdc++.h> using namespace std; const int maxn=1e7+…

GCD 题意:输入N,M(2<=N<=1000000000, 1<=M<=N), 设1<=X<=N,求使gcd(X,N)>=M的X的个数.  (文末有题) 知识点:   欧拉函数.http://www.cnblogs.com/shentr/p/5317442.html 题解一: 当M==1时,显然答案为N. 当M!=1.  X是N的因子的倍数是 gcd(X,N)>1 && X<=N 的充要条件.so  先把N素因子分解, N=     …

题意:给出N,求所有满足i<j<=N的gcd(i,j)之和 这题去年做过一次... 设f(n)=gcd(1,n)+gcd(2,n)+......+gcd(n-1,n),那么answer=S[N]=f(1)+f(2)+...+f(N). 先求出每一个f(n). 令g(n,i)=[满足gcd(x,n)=i且x<N的x的数量],i是n的约数 那么f(n)=sigma[i*g(n,i)] (i即gcd的值,g(n,i)为数量) 又注意到gcd(x,n)=i -> gcd(x/i,n/i)=…

GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4272    Accepted Submission(s): 1492 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 Again Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1997    Accepted Submission(s): 772 Problem Description Do you have spent some time to think and try to solve those unsolved problem aft…

Problem Description Goffi is doing his math homework and he finds an equality on his text book: gcd(n−a,n)×gcd(n−b,n)=nk. Goffi wants to know the number of (a,b) satisfy the equality, if n and k are given and 1≤a,b≤n. Note: gcd(a,b) means greatest co…

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) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the t…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1695 题意:x位于区间[a, b],y位于区间[c, d],求满足GCD(x, y) = k的(x, y)有多少组,不考虑顺序. 思路:a = c = 1简化了问题,原问题可以转化为在[1, b/k]和[1, d/k]这两个区间各取一个数,组成的数对是互质的数量,不考虑顺序.我们让d > b,我们枚举区间[1, d/k]的数i作为二元组的第二位,因为不考虑顺序我们考虑第一位的值时,只用考虑小于i的情…

GCD Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 76 Accepted Submission(s): 50   Problem Description The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b…