并不对劲的bzoj2820:p2257:YY的GCD

题目大意

\(t\)(\(t\leq10^4\))组数据，给定\(n,m\)(\(n,m\leq10^6\))求

\[\sum_{x=1}^{n}\sum_{y=1}^{m}[gcd(x,y)=1]
\]

题解

这个人（点这里）讲得很清楚\(\color{white}{\text{shing太强了}}\)

代码

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define dwn(i,x,y) for(register int i=(x);i>=(y);--i)
#define maxn 10000010
#define lim (maxn-10)
#define LL long long
using namespace std;
int read()
{
	int x=0,f=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-')ch=getchar();
	if(ch=='-')f=-1,ch=getchar();
	while(isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	return x*f;
}
void write(LL x)
{
	if(x==0){putchar('0'),putchar('\n');return;}
	int f=0;char ch[20];
	if(x<0)putchar('-'),x=-x;
	while(x)ch[++f]=x%10+'0',x/=10;
	while(f)putchar(ch[f--]);
	putchar('\n');
	return;
}
int t,n,m,p[maxn],no[maxn],mu[maxn],cnt;
LL f[maxn];
int main()
{
	no[1]=mu[1]=1;
	rep(i,2,lim)
	{
		if(!no[i])p[++cnt]=i,mu[i]=-1;
		for(int j=1;j<=cnt&&i*p[j]<=lim;j++)
		{
			no[i*p[j]]=1;
			if(i%p[j]==0){mu[i*p[j]]=0;break;}
			mu[i*p[j]]=-mu[i];
		}
	}
	rep(i,1,cnt)for(int j=p[i];j<=lim;j+=p[i])f[j]+=mu[j/p[i]];
	rep(i,1,lim)f[i]+=f[i-1];
	t=read();
	while(t--)
	{
		n=read(),m=read();
		if(n>m)swap(n,m);LL ans=0;
		for(int l=1,r=0;l<=n;l=r+1)r=min(n/(n/l),m/(m/l)),ans+=(LL)(n/l)*(LL)(m/l)*(f[r]-f[l-1]);
		write(ans);
	}
	return 0;
}