P3327 [SDOI2015]约数个数和

题意：

设 d(x) 为 x 的约数个数，给定 n,m，求
${\sum }_{i=1}^n{\sum }_{j=1}^{m}d(i,j)$

题解：

代码：

// Problem: P3327 [SDOI2015]约数个数和
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3327
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// By Jozky#include <bits/stdc++.h>
#include <unordered_map>
#define debug( a, b ) printf( "%s = %d\n", a, b );
using namespace std;
typedef long long          ll;
typedef unsigned long long ull;
typedef pair< int, int >   PII;
clock_t                    startTime, endTime;
// Fe~Jozky
const ll                                         INF_ll  = 1e18;
const int                                        INF_int = 0x3f3f3f3f;
void                                             read(){};
template < typename _Tp, typename... _Tps > void read( _Tp& x, _Tps&... Ar )
{x         = 0;char c    = getchar();bool flag = 0;while ( c < '0' || c > '9' )flag |= ( c == '-' ), c = getchar();while ( c >= '0' && c <= '9' )x = ( x << 3 ) + ( x << 1 ) + ( c ^ 48 ), c = getchar();if ( flag )x = -x;read( Ar... );
}
template < typename T > inline void write( T x )
{if ( x < 0 ){x = ~( x - 1 );putchar( '-' );}if ( x > 9 )write( x / 10 );putchar( x % 10 + '0' );
}
void rd_test()
{
#ifdef LOCALstartTime = clock();freopen( "in.txt", "r", stdin );
#endif
}
void Time_test()
{
#ifdef LOCALendTime = clock();printf( "\nRun Time:%lfs\n", ( double )( endTime - startTime ) / CLOCKS_PER_SEC );
#endif
}
const int N = 5e4 + 5;
int       tot, mu[ N ], p[ N ];
long long s[ N ];
bool      vis[ N ];void init()
{mu[ 1 ] = 1;for ( int i = 2; i <= 5e4; ++i ){if ( !vis[ i ] )p[ ++tot ] = i, mu[ i ] = -1;for ( int j = 1; j <= tot && i * p[ j ] <= 5e4; ++j ){vis[ i * p[ j ] ] = 1;if ( i % p[ j ] == 0 ){mu[ i * p[ j ] ] = 0;break;}else{mu[ i * p[ j ] ] = -mu[ i ];}}}for ( int i = 1; i <= 5e4; ++i )mu[ i ] += mu[ i - 1 ];for ( int x = 1; x <= 5e4; ++x ){long long res = 0;for ( int i = 1, j; i <= x; i = j + 1 )j = x / ( x / i ), res += 1LL * ( j - i + 1 ) * ( x / i );s[ x ] = res;}
}
int main()
{init();int T;for ( scanf( "%d", &T ); T--; ){int n, m;scanf( "%d%d", &n, &m );if ( n > m )swap( n, m );long long ans = 0;for ( int i = 1, j; i <= n; i = j + 1 ){j = min( n / ( n / i ), m / ( m / i ) );ans += 1LL * ( mu[ j ] - mu[ i - 1 ] ) * s[ n / i ] * s[ m / i ];}printf( "%lld\n", ans );}return 0;
}