这题呀，其实除了最后筛积性函数的时候比较困难，剩下的都是套路……
首先要想到的是所有满足条件的(mu(gcd(i, j))
eq 0)，然后就是暴推了。
首先得到的式子是这样的
[ans = sum x * mu(x) ^ 2 sum _ {d = 1} ^ {lfloor frac{n}{x}
floor} mu(d) * d ^ 2 * S(lfloor frac{n}{xd}
floor) * S(lfloor frac{m}{xd}
floor)
]
然后用(T = xd)替换
[ans = sum _ {T = 1} ^ {n} S(frac{n}{T}) * S(frac{m}{T}) * sum_{x | T, mu(x)
eq 0} x * mu(frac{T}{x}) * (frac{T}{x}) ^ 2
]
前面的可以数论分块，用狄利克雷卷积易证后面的是一个积性函数，所以可以(O(n))预处理。
至于怎么预处理我实在没推出来，各位可以看看路由器大佬的博客
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 4e6 + 5;
//const ll mod = (1 << 30);
inline ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(last == '-') ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
int v[maxn], prm[maxn];
ll f[maxn], sum[maxn];
void init()
{
f[1] = 1;
for(int i = 2; i < maxn; ++i)
{
if(!v[i]) v[i] = i, prm[++prm[0]] = i, f[i] = i - i * i;
for(int j = 1; j <= prm[0] && i * prm[j] < maxn; ++j)
{
v[i * prm[j]] = prm[j];
if(i % prm[j] == 0)
{
int x = i / prm[j];
if(x % prm[j]) f[i * prm[j]] = -prm[j] * prm[j] * prm[j] * f[x];
else f[i * prm[j]] = 0;
break;
}
else f[i * prm[j]] = f[i] * f[prm[j]];
}
}
for(int i = 1; i < maxn; ++i) sum[i] = sum[i - 1] + f[i];
}
ll s(ll n)
{
return n * (n + 1) / 2;
}
ll solve(int n, int m)
{
ll ret = 0;
int Min = min(n, m);
for(int l = 1, r; l <= Min; l = r + 1)
{
r = min(n / (n / l), m / (m / l));
ret = ret + (sum[r] - sum[l - 1]) * s(n / l) * s(m / l);
}
return ret;
}
int main()
{
init();
int T = read();
while(T--)
{
int n = read(), m = read();
write(solve(n, m) & ((1 << 30) - 1)), enter;
}
return 0;
}