题意:求(sumlimits_{i=1}^n sumlimits_{j=1}^m lcm(i,j) : gcd(i,j) 是sf 无平方因子数)
无平方因子数?搞一个(mu(gcd(i,j)))不就行了..不对不对有正负,是(mu^2)才行
套路推♂倒 (ノ*・ω・)ノ
[egin{align*}
sumlimits_{i=1}^n sumlimits_{j=1}^m frac{ij}{gcd(i,j)} mu(gcd(i,j))^2
&=sum_{d=1}^n d mu(d^2) sum_{i=1}^{frac{n}{d}} sum_{j=1}^{frac{m}{d}}ij[gcd(i,j)=1]\
&= sum_{D=1}^n Dsum_{d|D} mu(d)^2 mu(frac{D}{d})frac{D}{d} f(frac{n}{D}, frac{m}{D}) \
end{align*}
]
woc那是个smg,自己卷自己? $g(i) = i cdot ((mu cdot mu) * (mu cdot id))(i) $
如果我没猜错,点乘和卷积没有什么律吧
(g(1) = 1)
(g(p) = p*(1-p))
观察那堆(mu),分成的两个因子都有的话,相同的质数必须一边一个啊要不就是0没贡献了
考虑(p mid i),(i)中还至少有一个(p),我们记录最小质因子的次数判断一下(ip)的质因子多于2个(g(ip)=0)了,正好两个的话肯定是一面一个,结果就是(g(frac{i}{p})*(-p))啊...不对不对,前面的(i)让你吃了?应该是(*(-pcdot p^2))
貌似还有更科学的想法,i除掉p后两个就互质了...
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int N=4e6+5, INF=1e9, P = 1<<30;
#define pii pair<int, int>
#define MP make_pair
#define fir first
#define sec second
typedef long long ll;
inline int read(){
char c=getchar();int x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
}
int n, m;
int notp[N], p[N], cp[N]; ll g[N];
void sieve(int n) {
g[1] = 1;
for(int i=2; i<=n; i++) {
if(!notp[i]) p[++p[0]] = i, g[i] = i*(1-i), cp[i] = 1;
for(int j=1; j<=p[0] && i*p[j]<=n; j++) {
int t = i*p[j];
notp[t] = 1;
if(i%p[j] == 0) {
cp[t] = cp[i]+1;
if(cp[t] <= 2) g[t] = g[i/p[j]] * (-p[j] * p[j] * p[j]) %P;
else g[t] = 0;
break;
}
cp[t] = 1;
g[t] = g[i] * g[p[j]];
}
}
for(int i=1; i<=n; i++) (g[i] += g[i-1]) %=P;
}
inline ll f(ll n, ll m) {return ( n*(n+1)/2 %P) * ( m*(m+1)/2 %P)%P;}
ll cal(int n, int m) {
ll ans=0; int r;
for(int i=1; i<=n; i=r+1) {
r = min(n/(n/i), m/(m/i));
( ans += (g[r] - g[i-1]) * f(n/i, m/i) )%=P;
}
return (ans+P)%P;
}
int main() {
//freopen("in","r",stdin);
sieve(N-1);
int T=read();
while(T--) {
n=read(); m=read();
if(n>m) swap(n, m);
printf("%lld
", cal(n, m));
}
}