题目大意
有不超过(50000)个询问，每次询问有多少正整数对(x),(y)，满足(xleqslant a)，(y leqslant b)，并且(gcd(x,y)=c)。其中(a,b,cleqslant 50000)
解题思路
我们发现
[Ans=f(n)=sum_{x=1}^{a}sum_{y = 1}^{b}[gcd( i, j)=c]
]
当括号内表达式为真时，值为(1)，否则为(0)。
同时，我们设
[F(n)=sum_{n|d}f(d)=lfloorfrac{a}{n}
floorlfloorfrac{b}{n}
floor
]
由莫比乌斯反演，我们得
[f(d)=sum_{d|n}mu(frac{n}{d})F(n)=sum_{d|n}mu(frac{n}{d})lfloorfrac{a}{n}
floorlfloorfrac{b}{n}
floor=sum_{i=1}^{min(lfloorfrac{a}{n}
floor,lfloorfrac{b}{n}
floor)}mu(i)lfloorfrac{a}{ni}
floorlfloorfrac{b}{ni}
floor
]
到这里，我们通过预处理(mu)的前缀和和整除分块，就可以在(O(T*sqrt{n}))解决。
参考程序
// luogu-judger-enable-o2
#include <bits/stdc++.h>
#define LL long long
using namespace std;
const LL Maxn = 50010;
LL Mu[ Maxn ], Sum[ Maxn ];
int Vis[ Maxn ];
LL Num;
LL Prime[ Maxn ];
void Init() {
Mu[ 1 ] = 1;
for( LL i = 2; i <= Maxn; ++i ) {
if( !Vis[ i ] ) Mu[ i ] = -1, Prime[ ++Num ] = i;
for( LL j = 1; j <= Num && Prime[ j ] * i <= Maxn; ++j ) {
Vis[ Prime[ j ] * i ] = 1;
if( i % Prime[ j ] == 0 ) break;
Mu[ i * Prime[ j ] ] = -Mu[ i ];
}
}
for( LL i = 1; i <= Maxn; ++i ) Sum[ i ] = Sum[ i - 1 ] + Mu[ i ];
return;
}
void Work() {
LL a, b, c;
scanf( "%lld%lld%lld", &a, &b, &c );
if( a > b ) swap( a, b );
LL Ans = 0;
for( LL x = 1, y; x <= a / c; x = y + 1 ) {
y = min( ( a / c ) / ( ( a / c ) / x ), ( b / c ) / ( ( b / c ) / x ) );
Ans += a / c / x * ( b / c / x ) * ( Sum[ y ] - Sum[ x - 1 ] );
}
printf( "%lld
", Ans );
return;
}
int main() {
Init();
LL t; scanf( "%lld", &t );
for( ; t; --t ) Work();
return 0;
}