题解 P3455 【[POI2007]ZAP-Queries】

· · 题解

显然，题目求的是：

\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=d] \sum_{i=1}^{ \lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor \frac{m}{d} \rfloor}[gcd(i,j)=1]

有一个与莫比乌斯函数有关的性质：\sum_{d | n}\mu(d)=[n=1]

\sum_{i=1}^{ \lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor \frac{m}{d} \rfloor}\sum_{k|gcd(i,j)}\mu(k)

将 k 往前移，然后是一个非常经典的式子了

\sum_{k=1}^{min(\lfloor \frac{n}{d} \rfloor,\lfloor \frac{m}{d} \rfloor)}\mu(k)\lfloor \frac{n}{kd} \rfloor \lfloor \frac{m}{kd} \rfloor

用数论分块即可解决。

#include <cstdio>
#include <iostream>
using namespace std;

const int MAXN = 10000000;
int t , a , b , d;

int k , prime[ MAXN + 5 ] , mu[ MAXN + 5 ] , f[ MAXN + 5 ];
bool vis[ MAXN + 5 ];

void sieve( ) {
    mu[ 1 ] = 1;
    for( int i = 2 ; i <= MAXN ; i ++ ) {
        if( !vis[ i ] ) {
            prime[ ++ k ] = i;
            mu[ i ] = -1;
        }
        for( int j = 1 ; j <= k && 1ll * i * prime[ j ] <= MAXN ; j ++ ) {
            vis[ i * prime[ j ] ] = 1;
            if( i % prime[ j ] == 0 ) break;
            mu[ i * prime[ j ] ] = -mu[ i ];

        }
    }
    for( int i = 1 ; i <= MAXN ; i ++ )
        f[ i ] = f[ i - 1 ] + mu[ i ];
}

long long solve( int n , int m ) {
    long long Ans = 0;
    for( int l = 1 , r ; l <= min( n / d , m / d ) ; l = r + 1 ) {
        r = min( n / ( n / l ) , m / ( m / l ) );
        Ans += 1ll * ( n / l / d ) * ( m / l / d ) *  ( f[ r ] - f[ l - 1 ] );
    }
    return Ans;
}

int main( ) {
    sieve( );
    scanf("%d",&t);
    while( t -- ) {
        scanf("%d %d %d",&a,&b,&d);
        printf("%lld\n",solve( a , b ));
    }
    return 0;
}