Description
设(d(x))为(x)的约数个数,给定(N,M),求$$sum_{i=1}{N}sum_{j=1}{M}d(ij)$$。
Input
输入文件包含多组测试数据。
第一行,一个整数(T),表示测试数据的组数。
接下来的(T)行,每行两个整数(N,M)。
Output
(T)行,每行一个整数,表示你所求的答案。
Sample Input
2
7 4
5 6
Sample Output
110
121
HINT
(1 le N, M le 50000)
(1 le T le 50000)
这题有个很屌的结论:$$sum_{i=1}{N}sum_{j=1}{M}d(ij)=sum_{i=1}{N}sum_{j=1}{M}lfloorfrac{N}{i}
floorlfloorfrac{M}{j}
floorlbrack gcd(i,j)=1
brack$$
根据PoPoQQQ博客所说的,我们可以先证明这个式子的成立:$$d(nm)=sum_{i mid n}sum_{j mid m}lbrack gcd(i,j)=1
brack$$
我们可以证明一下:我们对每个质数(p)单独算贡献,设(n=n’ imes p^{k_{1}}),(m=m’ imes p^{k_{2}})。那么,该质数(p)对答案的贡献显然为(k_{1}+k_{2}+1)。于是我们考虑$$d(nm)=sum_{i mid n}sum_{j mid m}lbrack gcd(i,j)=1
brack$$这个式子,发现(p)对之有贡献的数对((i,j))仍然是$$(p{k_{1}},1),(p{k_{1}-1},1) cdots (1,1) cdots (1,p{k_{2}-1}),(1,p{k_{2}})$$这(k_{1}+k_{2}+1)个,因此得证。
代入得$$ sum_{n = 1}^{N}sum_{m = 1}^{M}d(nm) = sum_{n = 1}^{N}sum_{m = 1}^{M}sum_{i mid n} sum_{j mid m}[gcd(i,j)=1]$$
我们转变枚举量,先枚举(i,j)就有
于是$$sum_{i=1}{N}sum_{j=1}{M}lfloorfrac{N}{i}
floorlfloorfrac{M}{j}
floorlbrack gcd(i,j)=1
brack$$这个式子我们可以上反演了。
反演化为$$sum_{i=1}{N}sum_{j=1}{M}lfloorfrac{N}{i}
floorlfloorfrac{M}{j}
floor sum_{g mid i;g mid j} mu(g)$$
转而枚举(g),于是就可得到$$sum_{g=1}{N}mu(g)sum_{i=1}{lfloor frac{N}{g}
floor}sum_{j=1}^{lfloor frac{M}{g}
floor}lfloor frac{N}{ig}
floorlfloor frac{M}{jg}
floor$$
再化一下就可得到$$sum_{g=1}{N}mu(g)sum_{i=1}{lfloor frac{N}{g}
floor}lfloor frac{N}{ig}
floorsum_{j=1}^{lfloor frac{M}{g}
floor}lfloor frac{M}{jg}
floor$$
又有$$lfloor frac{N}{ab}
floor=lfloor frac{lfloor frac{N}{a}
floor}{b}
floor$$
于是我们发现(sum_{i=1}^{lfloor frac{N}{g}
floor}lfloor frac{N}{ig}
floor)只与(lfloor frac{N}{g}
floor)有关,我们可以(O(n sqrt{n}))预处理$$f_{x}=sum_{i=1}^{x}lfloor frac{x}{i}
floor$$
有了这个后再化简$$sum_{g=1}^{N}mu(g)f_{lfloor frac{N}{g}
floor}f_{lfloor frac{M}{g}
floor}$$就可在(O(sqrt{n}))分段求了。皆大欢喜。
#include<iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;
typedef long long ll;
#define maxn (50010)
int f[maxn],mu[maxn],prime[maxn],n,m,tot; bool exist[maxn];
inline int calc(int x)
{
int ret = 0;
for (int i = 1,last;i <= x;i = last+1)
{
last = min(x,x/(x/i));
ret += (x/i)*(last-i+1);
}
return ret;
}
inline void ready()
{
mu[1] = 1;
for (int i = 2;i <= 50000;++i)
{
if (!exist[i]) { prime[++tot] = i; mu[i] = -1; }
for (int j = 1;j <= tot&&prime[j]*i <= 50000;++j)
{
exist[i*prime[j]] = true;
if (i % prime[j] == 0) { mu[i*prime[j]] = 0; break; }
mu[i*prime[j]] = -mu[i];
}
}
for (int i = 1;i <= 50000;++i) mu[i] += mu[i-1],f[i] = calc(i);
}
inline ll work()
{
if (n > m) swap(n,m);
ll ret = 0;
for (int i = 1,last;i <= n;i = last+1)
{
last = min(n,min(n/(n/i),m/(m/i)));
ret += (ll)(mu[last]-mu[i-1])*((ll)f[n/i]*f[m/i]);
}
return ret;
}
int main()
{
freopen("3994.in","r",stdin);
freopen("3994.out","w",stdout);
ready();
int T; scanf("%d",&T);
while (T--) scanf("%d %d",&n,&m),printf("%lld
",work());
fclose(stdin); fclose(stdout);
return 0;
}