Code: #include<cstdio> using namespace std; const int maxn=4000005; const int R=4000002; const int N=4000002; long long sumv[maxn],f[maxn]; int phi[maxn],prime[maxn],vis[maxn]; void solve(){ phi[1]=1; int cnt=0; for(int i=2;i<=R;++i){ if(!vis[i])…
题意:给定一个数 n,问你0<= a <=n, 0 <= b <= n,有多少个不同的最简分数. 析:这是一个欧拉函数题,由于当时背不过模板,又不让看书,我就暴力了一下,竟然AC了,才2s,题目是给了3s,很明显是由前面递推,前面成立的,后面的也成立, 只要判定第 i 个有几个,再加前 i-1 个就好,第 i 个就是判断与第 i 个互质的数有多少,这就是欧拉函数了. 代码如下: 这是欧拉函数的. #pragma comment(linker, "/STACK:102400…
Given the value of N, you will have to find the value of G. The definition of G is given below:G =i<N∑i=1j∑≤Nj=i+1GCD(i, j)Here GCD(i, j) means the greatest common divisor of integer i and integer j.For those who have trouble understanding summation no…