几个有用的结论：
记狄利克雷卷积： $(f*g)(n)=\sum\limits_{d|n} f(d) * g(\frac{n}{d})$
则有重要结论:$\mu * 1 = [n=1]$与$\varphi *1 = id$与$\mu * id=\varphi$
若$F(n)=\sum\limits_{d|n} f(d)$
则$f(n)=\sum\limits_{d|n} \mu(d) * F(\frac{n}{d})$
bzoj 2190 仪仗队
题目大意：
求$n*n$的矩形中能从左下角被直接看到的点的个数
思路：
设左下角$(0,0)$ 则相当于求$\sum\limits_{i=1}^{n-1} \sum\limits_{j=1}^{n-1} [gcd(i,j)==1]+2$（$(1,0),(0,1)$
将式子转化为$1+ 2* \sum\limits_{i=1}^{n-1} \sum\limits_{j=1}^{i} [gcd(i,j)==1] $（由于有$(1,0),(0,1),(1,1)$三个特殊点存在
然后发现满足欧拉函数的定义，则所求为$1+2* \sum\limits_{i=1}^{n-1} \varphi(i) $ 筛出欧拉函数即可
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<cmath> 6 #include<algorithm> 7 #include<queue> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #define ll long long 12 #define db double 13 #define inf 2139062143 14 #define MAXN 50010 15 #define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i) 16 #define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i) 17 #define ren for(register int i=fst[x];i;i=nxt[i]) 18 #define pb(i,x) vec[i].push_back(x) 19 #define pls(a,b) (a+b)%MOD 20 #define mns(a,b) (a-b+MOD)%MOD 21 #define mul(a,b) (1LL*(a)*(b))%MOD 22 using namespace std; 23 inline int read() 24 { 25 int x=0,f=1;char ch=getchar(); 26 while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();} 27 while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();} 28 return x*f; 29 } 30 int n,tot,p[MAXN+100],ntp[MAXN+100],phi[MAXN+100]; 31 void mem() 32 { 33 phi[1]=1; 34 rep(i,2,MAXN) 35 { 36 if(!ntp[i]) p[++tot]=i,phi[i]=i-1; 37 rep(j,1,tot) if(i*p[j]>MAXN) break; 38 else {ntp[i*p[j]]=1;if(i%p[j]) phi[i*p[j]]=phi[i]*phi[p[j]];else {phi[i*p[j]]=phi[i]*p[j];break;}} 39 } 40 rep(i,2,n-1) phi[i]+=phi[i-1]; 41 } 42 int main() 43 { 44 n=read();mem();printf("%d\n",n>1?phi[n-1]<<1|1:0); 45 }