[SDOI2008]仪仗队

实际上就是求Gcd(i - 1, j - 1) == 1 的 (i, j) i >= 2, j >= 2 的个数加2
i - 1 和 j - 1 互质 那不就是sigma phi(i - 1)
最后答案*2+1即可

# include <stdio.h># include <stdlib.h># include <iostream># include <algorithm># include <string.h># define IL inline# define RG register# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;IL ll Read(){    RG char c = getchar(); RG ll x = 0, z = 1;    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + c - '0';    return x * z;}int n, ans;//Gcd(i - 1, j - 1) == 1//sigma phi(i - 1)IL int Phi(RG int x){    RG int cnt = x;    for(RG int i = 2; i <= x; i++){        if(x % i) continue;        while(!(x % i)) x /= i;        cnt = cnt - cnt / i;    }    if(x > 1) cnt = cnt - cnt / x;    return cnt;}int main(RG int argc, RG char* argv[]){    n = Read();    for(RG int i = 2; i <= n; i++) ans += Phi(i - 1);    printf("%d\n", ans * 2 + 1);    return 0;}