题目大意：
对于给定的$n,m(n,mleq10^7)$，求$为质数displaystylesum_{x=1}^nsum_{y=1}^m[gcd(x,y)为质数]$。
思路：
设$p=gcd(x,y),x=ap,y=bp$，则：
$$
egin{align*}
原式&=sum_{1leq pleqmin(n,m)且p为质数}sum_{a=1}^{lfloorfrac{n}{p}
floor}sum_{b=1}^{lfloorfrac{m}{p}
floor}[gcd(a,b)=1]\
&=sum_{1leq pleqmin(n,m)且p为质数}sum_{a=1}^{lfloorfrac{n}{p}
floor}sum_{b=1}^{lfloorfrac{m}{p}
floor}sum_{d|gcd(a,b)}mu(d)\
&=sum_{1leq pleqmin(n,m)且p为质数}sum_{p=1}^{lfloorfrac{min(n,m)}{p}
floor}mu(p)sum_{a=1}^{lfloorfrac{n}{dp}
floor}sum_{b=1}^{lfloorfrac{m}{dp}
floor}1\
&=sum_{1leq pleqmin(n,m)且p为质数}sum_{p=1}^{lfloorfrac{min(n,m)}{p}
floor}mu(p)lfloorfrac{n}{dp}
floorlfloorfrac{m}{dp}
floor
end{align*}
$$
令$k=dp$，则：
$$
egin{align*}
原式&=sum_{k=1}^{min(n,m)}sum_{p为质数且p|k}mu(p)lfloorfrac{n}{k}
floorlfloorfrac{m}{k}
floor
end{align*}
$$
其中$为质数且displaystylesum_{p为质数且p|k}mu(p)$可以预处理，$k$可以用数论分块枚举。
1 #include<cstdio> 2 #include<cctype> 3 #include<algorithm> 4 typedef long long int64; 5 inline int getint() { 6 register char ch; 7 while(!isdigit(ch=getchar())); 8 register int x=ch^'0'; 9 while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0'); 10 return x; 11 } 12 const int N=10000001,M=664580; 13 bool vis[N]; 14 int mu[N],p[M]; 15 int64 sum[N]; 16 inline void sieve() { 17 mu[1]=1; 18 for(register int i=2;i<N;i++) { 19 if(!vis[i]) { 20 p[++p[0]]=i; 21 mu[i]=-1; 22 } 23 for(register int j=1;j<=p[0]&&i*p[j]<N;j++) { 24 vis[i*p[j]]=true; 25 if(i%p[j]==0) { 26 mu[i*p[j]]=0; 27 break; 28 } 29 mu[i*p[j]]=-mu[i]; 30 } 31 } 32 for(register int i=1;i<=p[0];i++) { 33 for(register int j=1;j*p[i]<N;j++) { 34 sum[j*p[i]]+=mu[j]; 35 } 36 } 37 for(register int i=1;i<N;i++) { 38 sum[i]+=sum[i-1]; 39 } 40 } 41 int main() { 42 sieve(); 43 for(register int i=getint();i;i--) { 44 const int n=getint(),m=getint(),lim=std::min(n,m); 45 int64 ans=0; 46 for(register int i=1,j;i<=lim;i=j+1) { 47 j=std::min(n/(n/i),m/(m/i)); 48 ans+=(sum[j]-sum[i-1])*(n/i)*(m/i); 49 } 50 printf("%lld ",ans); 51 } 52 return 0; 53 }