题面
https://www.luogu.com.cn/problem/P1829
题解
前置知识
- 线性筛积性函数 https://www.cnblogs.com/zhoushuyu/p/8275530.html
- 莫比乌斯反演:《具体数学:计算机科学基础》4.9节
要求\({\sum_{i=1}^{n}}{\sum_{j=1}^{m}}[i,j]\)。
首先转化成为\({\sum_{i=1}^{n}}{\sum_{j=1}^{m}}{\frac{ij}{(i,j)}}\)。
然后设\(f(x) = {\frac{1}{x}}\),\(g = f {\times} {\mu}\)。(此处及以下“\({\times}\)”表示狄利克雷卷积)由莫比乌斯反演,有\(f=g{\times}I\)。
\[{\sum_{i=1}^{n}}{\sum_{j=1}^{m}}ijf((i,j))\]
\[={\sum_{i=1}^{n}}{\sum_{j=1}^{m}}ij{\sum_{d|(i,j)}}g(d)\]
\[={\sum_{i=1}^{n}}{\sum_{j=1}^{m}}ij{\sum_{d|i,d|j}}g(d)\]
\[={\sum_{d}}g(d){\sum_{i=1}^{n}}{\sum_{j=1}^{m}}[d|i][d|j]ij\]
\[={\sum_{d}}g(d)({\sum_{i=1}^{n}}[d|i]i)({\sum_{j=1}^{n}}[d|j]j)\]
注意到两个括号内分别是
\[d+2d+…+{\lfloor}{\frac{n}{d}}{\rfloor}d\]
\[d+2d+…+{\lfloor}{\frac{m}{d}}{\rfloor}d\]
这两个等差数列之和,于是原式=
\[{\sum_d}g(d)d^2({\frac{1}{2}}{\lfloor}{\frac{n}{d}}{\rfloor}({\lfloor}{\frac{n}{d}}{\rfloor}+1))({\frac{1}{2}}{\lfloor}{\frac{m}{d}}{\rfloor}({\lfloor}{\frac{m}{d}}{\rfloor}+1))\]
其中
\[g(d)={\sum_{p|d}}f({\frac{d}{p}}){\mu}(p)\]
\[={\sum_{p|d}}{\frac{p}{d}}{\mu(p)}\]
为了避免出现小数或者分数,简化计算,我们令\(h(d)=d*g(d)\)
那么有
\[h(d)={\sum_{p|d}}p{\mu(p)}\]
这是一个积性函数,可以线性求出。同时有原式=
\[{\sum_d}h(d)d({\frac{1}{2}}{\lfloor}{\frac{n}{d}}{\rfloor}({\lfloor}{\frac{n}{d}}{\rfloor}+1))({\frac{1}{2}}{\lfloor}{\frac{m}{d}}{\rfloor}({\lfloor}{\frac{m}{d}}{\rfloor}+1))\]
可以枚举d,O(min(n,m))求解。
代码
#include
using namespace std;
#define N 10000000
#define rg register
#define mod 20101009
#define F(x) (1ll*x*(x+1)/2%mod)
namespace ModCalc{
inline void Inc(int &x,int y){
x += y;if(x > mod)x -= mod;
}
inline void Dec(int &x,int y){
x -= y;if(x < 0)x += mod;
}
inline void Tms(int &x,int y){
x = 1ll * x * y % mod;
}
inline int Add(int x,int y){
Inc(x,y);return x;
}
inline int Sub(int x,int y){
Dec(x,y);return x;
}
inline int Mul(int x,int y){
Tms(x,y);return x;
}
}
using namespace ModCalc;
int pn;
int pri[1200000+5];
bool isp[N+5];
int h[N+5];
int P[N+5]; //P[i]表示i的素因数分解式中底数最小的那一项的大小
inline void Eular(){ //线性筛h[]
pn = 0;
h[1] = 1;
for(rg int i = 2;i <= N;i++)isp[i] = 1;
for(rg int i = 2;i <= N;i++){
if(isp[i]){
pri[++pn] = i;
P[i] = i;
h[i] = mod + 1 - i;
}
for(rg int j = 1;i * pri[j] <= N;j++){
isp[i*pri[j]] = 0;
if(i % pri[j]){
P[i*pri[j]] = pri[j];
h[i*pri[j]] = Mul(h[i],h[pri[j]]);
}
else{
int I = i * pri[j];
P[I] = Mul(P[i],pri[j]);
if(P[I] == I)h[I] = h[i];
else h[I] = Mul(h[I/P[I]],h[P[I]]);
break;
}
}
}
}
int n,m;
int main(){
Eular();
scanf("%d%d",&n,&m);
int ans = 0;
int M = min(n,m);
for(rg int d = 1;d <= M;d++)Inc(ans,1ll*h[d]*d%mod*F(n/d)%mod*F(m/d)%mod);
cout << ans << endl;
return 0;
}