P1829 [国家集训队]Crash的数字表格 / JZPTAB
题目
P1829 [国家集训队]Crash的数字表格 / JZPTAB
这题解法较多，都有值得学习的部分
解法一
\(Ans=\sum_{i=1}^{n}\sum_{j=1}^{m}\frac{ij}{gcd(i,j)}\)
思考把\(\sum_{d|n}\mu(d)=[n=1]\)带进去
\(Ans=\sum_{d=1}^{min(n,m)}\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=d]\frac{ij}{d}\)
\(Ans=\sum_{d=1}^{min(n,m)}d\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}[gcd(i,j)=1]ij\)
至此，我们把式变成：
\(Ans=\sum_{d=1}^{min(n,m)}d\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}\sum_{x|gcd(i,j)}\mu(x)ij\)
我们来枚举\(x\)消掉\(\sum_{x|gcd(i,j)}\)：
\(Ans=\sum_{d=1}^{min(n,m)}d\sum_{x=1}^{min(\lfloor\frac{n}{d}\rfloor,\lfloor\frac{m}{d}\rfloor)}\mu(x)\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}ij[x|gcd(i,j)]\)
对于判断\([x|gcd(i,j)]\)也要消掉：
\(Ans=\sum_{d=1}^{min(n,m)}d\sum_{x=1}^{min(\lfloor\frac{n}{d}\rfloor,\lfloor\frac{m}{d}\rfloor)}\mu(x)\sum_{xu=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{xv=1}^{\lfloor\frac{m}{d}\rfloor}x^2uv\)
最后要处理的式子：
\(Ans=\sum_{d=1}^{min(n,m)}d\sum_{x=1}^{min(\lfloor\frac{n}{d}\rfloor,\lfloor\frac{m}{d}\rfloor)}x^2\mu(x)(\sum_{u=1}^{\lfloor\frac{n}{dx}\rfloor}u)(\sum_{v=1}^{\lfloor\frac{m}{dx}\rfloor}v)\)
直接分块，另\((\sum_{u=1}^{\lfloor\frac{n}{dx}\rfloor}u)(\sum_{v=1}^{\lfloor\frac{m}{dx}\rfloor}v)\)相同
解法二
\(\begin{aligned} Ans &= \sum_{d=1}^n\sum_{i=1}^{n}\sum_{j=1}^{m}\frac{ij}{d}[gcd(i,j)=d] \\ &= \sum_{d=1}^nd\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor }\sum_{j=1}^{\lfloor \frac{m}{d} \rfloor }ij[gcd(i,j)==1] \\ & = \sum_{d=1}^nd\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor }\sum_{j=1}^{\lfloor \frac{m}{d} \rfloor }ij\sum_{t|gcd(i,j)}\mu(t) \\ & = \sum_{d=1}^nd\sum_{t=1}^{\lfloor \frac{n}{d} \rfloor}\mu(t)\sum_{i=1}^{\frac{n}{dt}}it\sum_{j=1}^{\frac{m}{dt}}jt \\ & = \sum_{d=1}^nd\sum_{t=1}^{\lfloor \frac{n}{d} \rfloor}t^2\mu(t)\sum_{i=1}^{\frac{n}{dt}}i\sum_{j=1}^{\frac{m}{dt}}j \\ \end{aligned}\)
由于\(\sum_{i=1}^{\frac{n}{dt}}i\sum_{j=1}^{\frac{m}{dt}}j\)可以分块，所以枚举\(dt\)
\(\sum_{T=1}^n \sum_{i=1}^{\frac{n}{T}}i\sum_{j=1}^{\frac{m}{T}} \rfloor)T\sum_{t|T}t\mu(t)\)
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const LL p=20101009;
const int maxn=1e7+10;
inline int Read(){
LL x=0,f=1; char c=getchar();
while(c<'0'||c>'9'){
if(c=='-') f=-1; c=getchar();
}
while(c>='0'&&c<='9'){
x=(x<<3)+(x<<1)+c-'0',c=getchar();
}
return x*f;
}
int n,m;
LL ans;
int mu[maxn],prime[maxn];
LL sum[maxn];
bool visit[maxn];
inline void F_phi(LL max_n){
mu[1]=1;
int tot=0;
for(int i=2;i<=max_n;++i){
if(!visit[i]){
prime[++tot]=i,
mu[i]=-1;
}
for(int j=1;j<=tot&&i*prime[j]<=max_n;++j){
visit[i*prime[j]]=true;
if(i%prime[j]==0)
break;
else
mu[i*prime[j]]=-mu[i];
}
}
for(int i=1;i<=max_n;++i)
sum[i]=(sum[i-1]+(LL)mu[i]*i%p*i%p)%p;
}
int main(){
scanf("%d%d",&n,&m);
int N=min(n,m);
F_phi(N);
for(int d=1;d<=N;++d){
int maxx=n/d,maxy=m/d;
int x=min(maxx,maxy);
LL num=0;
for(int l=1,r;l<=x;l=r+1){
r=min(maxx/(maxx/l),maxy/(maxy/l));
num=(num+
((sum[r]-sum[l-1]+p)%p)*
((((1ll+maxx/l)%p)*1ll*(maxx/l)/2%p)%p)%p*
((((1ll+maxy/l)%p)*1ll*(maxy/l)/2%p)%p)%p)%p;
}
ans=(ans+(num*1ll*d)%p)%p;
}
printf("%lld",ans);
return 0;
}/*
1000000 1000000
9002207
*/
