2021-08-20P3327 [SDOI2015]约数个数和

Posted 2021-09-08 Jozky86

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 LOCAL
    startTime = clock();
    freopen( "in.txt", "r", stdin );
#endif
}
void Time_test()
{
#ifdef LOCAL
    endTime = 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;
}