# BZOJ 2301: [HAOI2011]Problem b

/*
BZOJ 2301: [HAOI2011]Problem b

莫比乌斯反演 + 容斥
将k除下来后就变为了一道原题

后像求二维前缀和一样减去重复计算的，再加上多减的即可
*/
#include <cstdio>
#include <iostream>

#define rg register
#define Max 50007
int p[Max], mu[Max], sm[Max]; bool is[Max];
typedef long long LL;
inline LL min (LL a, LL b) { return a < b ? a : b; }
{
rg char c = getchar ();
for (n = 0; !isdigit (c); c = getchar ());
for (; isdigit (c); n = n * 10 + c - '0', c = getchar ());
}

void Euler (int N)
{
rg int i, j, k; int C = 0; mu[1] = 1;
for (i = 2; i <= N; ++ i)
{
if (!is[i]) p[++ C] = i, mu[i] = -1;
for (j = 1; j <= C; ++ j)
{
if ((k = i * p[j]) > N) break;
is[k] = true;
if (i % p[j] == 0) { mu[k] = 0; break; }
else mu[k] = -mu[i];
}
}
for (i = 1; i <= N; ++ i) mu[i] += mu[i - 1];
}
LL Z (LL N, LL M)
{
LL res = 0; rg LL i, j;
if (N > M) std :: swap (N, M);
for (i = 1; i <= N; i = j + 1)
{
j = min (N / (N / i), M / (M / i));
res += (LL) (mu[j] - mu[i - 1]) * (N / i) * (M / i);
}
return res;
}
int main (int argc, char *argv[])
{
int T, A, B, C, D, K; read (T); rg int i;
Euler (Max - 1); LL Answer = 0;
for (; T; -- T)
{
}