链接
https://ac.nowcoder.com/acm/contest/549/J
题意
$\sum_{i=1}^n\sum_{j=1}^m\gcd(i,j)^2$ $(n,m\le 1e6)$
思路一
$\begin{aligned}
& \sum_{i=1}^n\sum_{j=1}^m\gcd(i,j)^2 \
= & \sum_{d=1}d^2\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=d] \
\end{aligned}$
记 $f(d)=\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=d]$
记 $F(t)=\sum_{d\mid t}f(d)=\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor$
$f(d)=\sum_{d\mid t}\mu(\frac{t}{d})F(t)=\sum_{d\mid t}\mu(\frac{t}{d})\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor$
枚举 $d$ 的倍数 $O(nlogn)$ 求解。
代码一
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
| #include <bits/stdc++.h>
#define SZ(x) (int)(x).size()
#define ALL(x) (x).begin(),(x).end()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define FI first
#define SE second
using namespace std;
typedef double DB;
typedef long double LD;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef vector<PII> VPII;
const LL MOD=1e9+7;
const int N=1e6+5;
int mn[N],mu[N];
VI p;
void get_mu(int n) {
mu[1]=1;
for(int i=2;i<=n;i++) {
if(!mn[i]) mn[i]=i,mu[i]=-1,p.PB(i);
for(auto x:p) {
if(x*i>n) break;
mn[x*i]=x;
if(i%x==0) {mu[x*i]=0;break;}
mu[x*i]=-mu[i];
}
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n,m;cin>>n>>m;
get_mu(1e6);
LL res=0;
int lim=min(n,m);
for(int i=1;i<=lim;i++) {
for(int j=i;j<=lim;j+=i) {
res=(res+1ll*i*i%MOD*mu[j/i]*(n/j)%MOD*(m/j)%MOD)%MOD;
}
}
cout<<res<<'\n';
return 0;
}
|
思路二
$\begin{aligned}
& \sum_{i=1}^n\sum_{j=1}^m\gcd(i,j)^2 \
= & \sum_{d=1}d^2\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=d] \
= & \sum_{d=1}d^2\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}[\gcd(i,j)=1] \
\end{aligned}$
记 $g(n,m)=\sum_{i=1}^{n}\sum_{j=1}^{m}[\gcd(i,j)=1]=\sum_{d=1}\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor$
记 $f(n,m)=\sum_{d=1}d^2g(\lfloor\frac{n}{d}\rfloor,\lfloor\frac{m}{d}\rfloor)$
做两次数论分块。
代码二
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
| #include <bits/stdc++.h>
#define SZ(x) (int)(x).size()
#define ALL(x) (x).begin(),(x).end()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define FI first
#define SE second
using namespace std;
typedef double DB;
typedef long double LD;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef vector<PII> VPII;
const LL MOD=1e9+7;
const int N=1e6+5;
int mn[N];
LL mu[N],sum[N];
VI p;
void get_mu(int n) {
mu[1]=1;
for(int i=2;i<=n;i++) {
if(!mn[i]) mn[i]=i,mu[i]=-1,p.PB(i);
for(auto x:p) {
if(x*i>n) break;
mn[x*i]=x;
if(i%x==0) {mu[x*i]=0;break;}
mu[x*i]=-mu[i];
}
}
for(int i=1;i<=n;i++) mu[i]=(mu[i-1]+MOD+mu[i])%MOD;
for(int i=1;i<=n;i++) sum[i]=(sum[i-1]+1ll*i*i%MOD)%MOD;
}
LL g(int n,int m) {
int lim=min(n,m);
LL ret=0;
for(int l=1,r;l<=lim;l=r+1) {
r=min(n/(n/l),m/(m/l));
ret=(ret+(mu[r]+MOD-mu[l-1])%MOD*(n/l)%MOD*(m/l)%MOD)%MOD;
}
return ret;
}
LL f(int n,int m) {
int lim=min(n,m);
LL ret=0;
for(int l=1,r;l<=lim;l=r+1) {
r=min(n/(n/l),m/(m/l));
ret=(ret+(sum[r]+MOD-sum[l-1])%MOD*g(n/l,m/l)%MOD)%MOD;
}
return ret;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n,m;cin>>n>>m;
get_mu(1e6);
cout<<f(n,m)<<'\n';
return 0;
}
|