BZOJ-1257、2301、1101几道数论题

题目：

http://www.lydsy.com/JudgeOnline/problem.php?id=1257

http://www.lydsy.com/JudgeOnline/problem.php?id=2301

http://www.lydsy.com/JudgeOnline/problem.php?id=1101

数论一向就是本蒟蒻最渣渣的内容，无奈去啃了一下：

1257: [CQOI2007]余数之和sum:

由于a%b=a-(a/b)*b,那么对于a/b进行分段统计即可，复杂度sqrt(n)

Code:

#include 

 

using namespace std ;

 

#define ll long long

 

ll n , k , ans = 0 ;

 

int main(  ) {

    cin >> n >> k ;

    for ( ll i = 1 ; i <= n ; ) {

        if ( i > k ) {

            ans += ( n - i + 1 ) * k ; break ;

        }

        ll pos = min( k / ( k / i ) , n ) ;

        ans += ( k * ( pos - i + 1 ) - ( k / i ) * ( pos - i + 1 ) * ( i + pos ) / 2 ) ;

        i = pos + 1 ;

    }

    cout << ans << endl ;

    return 0 ;

}

1101: [POI2007]Zap：

要求1<=x<=a,1<=y<=b,gcd(x,y)=c的数对数目，那么就相当于求，gcd(x/c,y/c)=1,1<=x<=a/c,1<=y<=b/c的数目，然后再用神奇的莫比乌斯反演可以得到答案是，sigma u(d)(a/(cd))(b/(cd)),然后就像上一题一样，对a/(cd)分段统计即可，总复杂度O(n^1.5)

Code:

#include 

#include 

#include 

 

using namespace std ;

 

#define MAX 50100

#define Sum( l , r ) ( sum[ r ] - sum[ l - 1 ] )

 

bool f[ MAX ] ;

int u[ MAX ] , sum[ MAX ] , maxd = 0 , n , A[ MAX ] , B[ MAX ] , C[ MAX ] ;

 

void Init(  ) {

    scanf( "%d" , &n ) ;

    for ( int i = 0 ; i ++ < n ; ) {

        scanf( "%d%d%d" , &A[ i ] , &B[ i ] , &C[ i ] ) ;

        maxd = max( maxd , max( A[ i ] , B[ i ] ) / C[ i ] ) ;

    }

    memset( f , true , sizeof( f ) ) ;

    f[ 0 ] = f[ 1 ] = false ;

    memset( u , 0 , sizeof( u ) ) ; u[ 1 ] = 1 ;

    for ( int i = 1 ; i ++ < maxd ; ) if ( f[ i ] ) {

        u[ i ] = - 1 ;

        for ( int j = i << 1 ; j <= maxd ; j += i ) {

            if ( ! ( ( j / i ) % i ) ) u[ j ] = 0 ;

            else u[ j ] = u[ j / i ] * - 1 ;

            f[ j ] = false ;

        }

    }

    sum[ 0 ] = 0 ;

    for ( int i = 1 ; i <= maxd ; ++ i ) sum[ i ] = sum[ i - 1 ] + u[ i ] ;

}

 

void Solve(  ) {

    for ( int j = 0 ; j ++ < n ; ) {

        int a , b , c , ans = 0 ;

        a = A[ j ] , b = B[ j ] , c = C[ j ] ;

        a /= c , b /= c ;

        if ( a > b ) swap( a , b ) ;

        for ( int i = 1 ; i <= a ; ) {

            int pos = min( a / ( a / i ) , b / ( b / i ) ) ;

            ans += Sum( i , pos ) * ( a / i ) * ( b / i ) ;

            i = pos + 1 ;

        }

        printf( "%d\n" , ans ) ;

    }

}

 

int main(  ) {

    Init(  ) ;

    Solve(  ) ;

    return 0 ;

}

2301: [HAOI2011]Problem b：

上一道题的略略加强版，设f(a,b)为gcd(x,y)=1,1<=x<=a,1<=y<=b的数目，那么gcd(x,y)=1,a<=x<=b,c<=y<=d的数目用容斥定理弄一下就成了f(b,d)-f(a-1,d)-f(b,c-1)+f(a-1,c-1)，然后用上题的方法就可以统计出来了。

Code:

#include 

#include 

#include 

 

using namespace std ;

 

#define Sum( l , r ) ( sum[ r ] - sum[ l - 1 ] )

#define MAX 51000

 

int n , A[ MAX ] , B[ MAX ] , C[ MAX ] , D[ MAX ] , K[ MAX ] ;

int u[ MAX ] , sum[ MAX ] , maxd = 0 ;

bool f[ MAX ] ;

 

void Init(  ) {

    scanf( "%d" , &n ) ;

    for ( int i = 0 ; i ++ < n ; ) {

        scanf( "%d%d%d%d%d" , A + i , B + i , C + i , D + i , K + i ) ;

        maxd = max( maxd , max( B[ i ] , D[ i ] ) / K[ i ] ) ;

    }

    memset( f , true , sizeof( f ) ) ;

    memset( u , 0 , sizeof( u ) ) ;

    memset( sum , 0 , sizeof( sum ) ) ;

    f[ 0 ] = f[ 1 ] = false , u[ 1 ] = 1 ;

    for ( int i = 1 ; i ++ < maxd ; ) if ( f[ i ] ) {

        u[ i ] = - 1 ;

        for ( int j = i << 1 ; j <= maxd ; j += i ) {

            if ( ! ( ( j / i ) % i ) ) u[ j ] = 0 ; else u[ j ] = u[ j / i ] * - 1 ;

            f[ j ] = false ;

        }

    }

    for ( int i = 0 ; i ++ < maxd ; ) sum[ i ] = sum[ i - 1 ] + u[ i ] ;

}

 

int query( int a , int b ) {

    int ans = 0 ;

    for ( int i = 1 ; i <= min( a , b ) ; ) {

        int pos = min( a / ( a / i ) , b / ( b / i ) ) ;

        ans += Sum( i , pos ) * ( a / i ) * ( b / i ) ;

        i = pos + 1 ;

    }

    return ans ;

}

 

void Solve(  ) {

    for ( int i = 0 ; i ++ < n ; ) {

        A[ i ] -- , C[ i ] -- ;

        A[ i ] /= K[ i ] , B[ i ] /= K[ i ] , C[ i ] /= K[ i ] , D[ i ] /= K[ i ] ;

        printf( "%d\n" , query( B[ i ] , D[ i ] ) - query( A[ i ] , D[ i ] ) - query( B[ i ] , C[ i ] ) + query( A[ i ] , C[ i ] ) ) ;

    }

}

 

int main(  ) {

    Init(  ) ;

    Solve(  ) ;

    return 0 ;

}