xiayuyang

题解好像有一些typo，不过问题不大。本题就是经典莫比乌斯反演套路题关键问题是读入a的时候要把aaa转化成f(a)f(a)f(a)具体的式子以后来补#include<bits/stdc++.h>using namespace std;#define int long longconst int mod=1e9+7;const int maxn=2e5+5,N=2e5;int mu[maxn],p[maxn],cnt,vis[maxn],f[maxn],b[maxn],n,mx;

题目链接解法首先观察一下，可以发现对于a,b，正方形的边长是lcm(a,b)。那么买的地板的数量是(a∗b)gcd(a,b)2\frac{(a*b)}{gcd(a,b)^2}gcd(a,b)2(a∗b)​那么对于一个n的答案就是：ans=∏i=1n∏j=1ni∗jgcd(i,j)2ans=\prod_{i=1}^n \prod_{j=1}^n \frac{i*j}{gcd(i,j)^2}an...

题目链接题意给两个数n,k问对于一个长度为n,元素范围为1到k1到k1到k的数组a,有多少个a的总gcd为1,记这个值为bib_ibi​,要求输出∑i=1k(bi xor i)\sum_{i=1}^k(b_i ~xor ~i)∑i=1k​(bi​ xor i)数据范围n,k≤2e6n,k\le 2e6n,k≤2e6解法首先列出bbb的式子:bm...

题意n<=1e9n<=1e9n<=1e9,答案对2322^{32}232取膜部分分n<=1e6n<=1e6n<=1e6,n<=1e7n<=1e7n<=1e7首先发现mu(i∗j)=mu(i)∗mu(j)∗[gcd(i,j)==1]mu(i*j)=mu(i)*mu(j)*[gcd(i,j)==1]mu(i∗j)=mu(i)∗mu(j)∗[...

题意现在有TTT个询问,每个询问给你三个整数n,m,an,m,an,m,a,求对于所有1<=i<=n,1<=j<=m,1<=gcd(i,j)<=a1<=i<=n,1<=j<=m,1<=gcd(i,j)<=a1<=i<=n,1<=j<=m,1<=gcd(i,j)<=a的i,ji,ji,j的...