bzoj3994:[SDOI2015]约数个数和
题目链接
题解
结论与结论的证明参考了rqy的博客
计算\(d(ij)\)时，把ij的每个约数d映射到\((a=gcd(d,i),b=(gcd(d,i),\frac{d}{gcd(d, i)})\)
可知\(a,b\)分别是\(i,j\)的因数，且\((a,b)\)对应一个因数当且仅当\(gcd(\frac ia, b) = 1\)，所以
\[d(ij) = \sum_{x|i}\sum_{y|j} [gcd(\frac ix, y) = 1] = \sum_{x'|i}\sum_{y|j} [gcd(x', y) = 1]
\]
有了这个式子之后
\[\sum_{i=1}^n\sum_{j=1}^m d(ij) =\sum_{i=1}^n\sum_{j=1}^m\sum_{x|i}\sum_{y|j} [gcd(x, y) = 1]$$ $$=\sum_{i=1}^n\sum_{j=1}^m \left\lfloor\frac ni\right\rfloor \left\lfloor\frac mj\right\rfloor [gcd(i, j) = 1]$$ $$=\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d|i}\sum_{d|j}\mu(d)\lfloor\frac{n}{i}\rfloor\lfloor\frac{m}{j}\rfloor$$ $$ =\sum_{d=1}^{min(n,m)}\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}\mu(d)\lfloor\frac{n}{i}\rfloor\lfloor\frac{m}{j}\rfloor $$
\]
=\sum_{d=1}{min(n,m)}\mu(d)\sum_{i=1}\rfloor}\lfloor\frac{n}{i}\rfloor\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}\lfloor\frac{m}{j}\rfloor$$
用\(f(x)=\sum_{i=1}^{x}\lfloor\frac{x}{i}\rfloor\)表示1-x的约数个数和
\[\sum_{d=1}^{min(n,m)}\mu(d)f(\lfloor\frac{n}{d}\rfloor)f(\lfloor\frac{m}{d}\rfloor)$$预处理$\mu(n)$前缀和
对于$f$我们可以$O(n\sqrt n)$预处理,查询为$O(\sqrt n)$
##代码
```c++
#include<cstdio>
#include<cstring>
#include<algorithm>
const int maxn = 50007;
#define int long long
inline int read() {
int x=0,f=1;
char c=getchar();
while(c<'0'||c>'9'){if(c=='0')f=-1;c=getchar();}
while(c<='9'&&c>='0')x=x*10+c-'0',c=getchar();
return x*f;
}
int f[maxn],mu[maxn],num,prime[maxn];
bool vis[maxn];
inline int countf(int x) {
int ans=0;
for(int i=1,last;i<=x;i=last+1) {
last=x/(x/i);
ans+=x/i*(last-i+1);
}
return ans;
}
void get_mu() {
mu[1]=1;
for(int i=2;i<maxn;++i) {
if(!vis[i])prime[++num]=i,mu[i]=-1;
for(int j=1;j<=num&&i*prime[j]<maxn;++j) {
vis[prime[j]*i]=1;
if(i%prime[j]==0)mu[prime[j]*i]=0;
else mu[i*prime[j]]=-mu[i];
}
}
for(int i=1;i<maxn;++i)mu[i]+=mu[i-1];
}
void init() {
for(int i=1;i<maxn;++i)f[i]=countf(i);
get_mu();
}
int solve(int n,int m) {
int ans=0;
for(int i=1,last;i<=std::min(n,m);i=last+1) {
last=std::min(n/(n/i),m/(m/i));
ans+=(mu[last]-mu[i-1])*f[n/i]*f[m/i];
}
return ans;
}
main() {
init();
int T=read();
for(;T--;) {
printf("%lld\n",solve(read(),read()));
}
int tmp=0;
return 0;
}
```\]
