2015
题意：\(d(i)\)为i的约数个数，求\(\sum\limits_{i=1}^n \sum\limits_{j=1}^m d(ij)\)
\(ij\)都爆int了....
一开始想容斥一下用\(d(i)\)和\(d(j)\)算\(d(ij)\)，发现不行...
然后翻题解看到了一步好神的转化：
\[d(nm) = \sum_{i\mid n} \sum_{j\mid m} [gcd(i,j)=1]
\]
\]
晚上再补吧还是没拿草稿纸...
补：
\(Proof.\)
- 首先注意约数个数 相同的算一个
约数个数公式\(\prod (a_i+1)\)
考虑一个质因子，\(p^x,p^y \rightarrow p^x p^y\)
\(x+y+1\)对应了\(gcd(p^x, 1)...gcd(1, 1)...gcd(1,p^y)\)
质因子相互独立，乘起来
然后愉♂悦的套路推♂倒
\[\begin{align*}
\sum_{i=1}^n \sum_{j=1}^m d(ij) &= \sum_{i=1}^n \sum_{j=1}^m \sum_{x\mid i} \sum_{y\mid j} [gcd(x,y)=1]\\
先枚举约数,交换i,j\ x,y\\
&=\sum_{i=1}^n \sum_{j=1}^m \sum_{d\mid i,d\mid j}\mu(d) \frac{n}{i} \frac{m}{i}\\
&=\sum_{d=1}^n \mu(d)\sum_{i=1}^\frac{n}{i} \sum_{j=1}^\frac{m}{i} \frac{n}{id}\frac{m}{jd}\\
&=\sum_{d=1}^n \mu(d) f(\frac{n}{id})f(\frac{m}{jd})\\
\end{align*}
\]
\sum_{i=1}^n \sum_{j=1}^m d(ij) &= \sum_{i=1}^n \sum_{j=1}^m \sum_{x\mid i} \sum_{y\mid j} [gcd(x,y)=1]\\
先枚举约数,交换i,j\ x,y\\
&=\sum_{i=1}^n \sum_{j=1}^m \sum_{d\mid i,d\mid j}\mu(d) \frac{n}{i} \frac{m}{i}\\
&=\sum_{d=1}^n \mu(d)\sum_{i=1}^\frac{n}{i} \sum_{j=1}^\frac{m}{i} \frac{n}{id}\frac{m}{jd}\\
&=\sum_{d=1}^n \mu(d) f(\frac{n}{id})f(\frac{m}{jd})\\
\end{align*}
\]
问题就是\(f(n)=\sum_{i=1}^n\frac{n}{i}\)怎么求了
可以n根n预处理...
更巧妙的做法是，发现\(f\)就是\(d\)的前缀和，因为\(\frac{n}{i}\)表示的就是\(1..n\)有几个i的倍数
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=5e4+5;
typedef long long ll;
#define pii pair<int, int>
#define MP make_pair
#define fir first
#define sec second
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], mu[N]; ll f[N]; pii lp[N];
void sieve(int n) {
mu[1] = 1; f[1] = 1;
for(int i=2; i<=n; i++) {
if(!notp[i]) p[++p[0]] = i, mu[i] = -1, f[i] = 2, lp[i] = MP(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) {
mu[t] = 0;
lp[t] = MP(p[j], lp[i].sec + 1);
f[t] = f[i] / (lp[i].sec + 1) * (lp[t].sec + 1);
break;
}
mu[t] = -mu[i];
lp[t] = MP(p[j], 1);
f[t] = f[i] * (lp[t].sec + 1);
}
}
for(int i=1; i<=n; i++) mu[i] += mu[i-1], f[i] += f[i-1];
}
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 += (mu[r] - mu[i-1]) * f[n/i] * f[m/i];
}
return ans;
}
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\n",cal(n, m));
}
}