【HDU】5321 Beautiful Set【枚举k求贡献，欧拉函数应用】

来源：互联网发布：java图形界面代码编辑：程序博客网时间：2024/05/29 17:05
传送门：【HDU】5321 Beautiful Set
my code:
#include <stdio.h>#include <string.h>#include <vector>#include <algorithm>using namespace std ;typedef long long LL ;#define clr( a , x ) memset ( a , x , sizeof a )const int MAXN = 100005 ;const int mod = 258280327 ;bool prime[MAXN] ;int phi[MAXN] ;int cnt[MAXN] ;int id[MAXN] ;int f[MAXN] ;int vf[MAXN] ;int n ;void exgcd ( int a , int b , int& x , int& y ) {    if ( b ) {        exgcd ( b , a % b , y , x ) ;        y -= a / b * x ;    } else x = 1 , y = 0 ;}int inv ( int a ) {    int x , y , b = mod ;    exgcd ( a , b , x , y ) ;    if ( x < 0 ) x += mod ;    return x ;}int cmp ( int a , int b ) {    return cnt[a] > cnt[b] ;}void calc () {    for ( int i = 1 ; i < MAXN ; ++ i ) {        id[i] = i ;        for ( int j = i + i ; j < MAXN ; j += i ) {            cnt[i] += cnt[j] ;        }    }    sort ( id + 1 , id + MAXN , cmp ) ;}void preprocess () {    f[0] = vf[0] = 1 ;    for ( int i = 1 ; i < MAXN ; ++ i ) {        phi[i] = i ;        f[i] = ( LL ) i * f[i - 1] % mod ;        vf[i] = inv ( f[i] ) ;    }    for ( int i = 2 ; i < MAXN ; ++ i ) if ( !prime[i] ) {        for ( int j = i ; j < MAXN ; j += i ) {            phi[j] = phi[j] / i * ( i - 1 ) ;            prime[j] = 1 ;        }    }}int c ( int a , int b ) {    return ( LL ) f[a] * vf[b] % mod * vf[a - b] % mod ;}void solve () {    int x ;    int ans1 = 0 , ans2 = 0 ;    clr ( cnt , 0 ) ;    for ( int i = 1 ; i <= n ; ++ i ) {        scanf ( "%d" , &x ) ;        cnt[x] ++ ;    }    calc () ;    for ( int i = 1 ; i < MAXN ; ++ i ) {        int tmp = 0 ;        for ( int j = 1 ; j < MAXN ; ++ j ) {            int idx = id[j] ;            if ( cnt[idx] < i ) break ;            int t = ( LL ) c ( cnt[idx] , i ) * phi[idx] % mod ;            tmp = ( tmp + t ) % mod ;        }        ans2 = ( ans2 + ( LL ) i * tmp ) % mod ;        tmp = ( LL ) tmp * f[i] % mod * f[n - i + 1] % mod ;        ans1 = ( ans1 + tmp ) % mod ;    }    if ( ans1 > ans2 ) printf ( "Mr. Zstu %d\n" , ans1 ) ;    else if ( ans1 < ans2 ) printf ( "Mr. Hdu %d\n" , ans2 ) ;    else printf ( "Equal %d\n" , ans1 ) ;}int main () {    preprocess () ;    while ( ~scanf ( "%d" , &n ) ) solve () ;    return 0 ;}
0 0