HDU4609 3-idiots

Posted 2022-09-23 chy-2003

题目链接

问题分析

题意就是求随机取\(3\)条边能组成长方形的概率。其实可以看做求方案数。

不妨将边长从小到大排序，得到\(a_1,a_2,\cdots,a_n\)。考虑枚举三角形边长中最大的一条\(a_i\)，那么\(a_i\)作为最大边时贡献就是\(a_1,a_2,\cdots,a_i-1\)中两项加起来大于\(a_i\)的方案数。然而发现这样比较难将复杂度降到\(O(n\log n)\)。

不妨换一个角度考虑。首先求出取两条边可能的所有情况，然后再枚举第三条边。和上面一样，让枚举的第三条边是三角形中最大的边，然后去重即可。由于边长范围只有\(1e5\)，通过母函数和FFT来求得。具体细节可以看代码。假装100组数据\(O(n\log n)\)能过就好了

注意使用long long

FFT不要忘记最后除以n

参考程序

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const long long MaxAlpha = 300010;
const long long FullAlpha = 100000;
const long long Maxn = 100010;
const double Pi = 3.14159265358979323846264;
void Work();
void Init();
int main() 
    int TestCases;
    scanf( "%d", &TestCases );
    for( ; TestCases--; ) Work(); 
    return 0;


long long n, m, A[ Maxn ];
long long Sum[ MaxAlpha ];
long long UpTo;

struct MyComplex 
    double Real, Imagine;
    MyComplex() 
    MyComplex( double _Real, double _Imagine ) : Real( _Real ), Imagine( _Imagine ) 
    inline MyComplex operator + ( const MyComplex Other ) const 
        return ( MyComplex )  Real + Other.Real, Imagine + Other.Imagine ;
    
    inline MyComplex operator - ( const MyComplex Other ) const 
        return ( MyComplex )  Real - Other.Real, Imagine - Other.Imagine ;
    
    inline MyComplex operator * ( const MyComplex Other ) const 
        return ( MyComplex )  Real * Other.Real - Imagine * Other.Imagine, Real * Other.Imagine + Imagine * Other.Real ;
    
;
MyComplex FFT_A[ MaxAlpha ], FFT_B[ MaxAlpha ], Omega[ MaxAlpha ];
long long FFT_Index[ MaxAlpha ];
void FFT( MyComplex *A, long long Len ) 
    for( long long i = 0; i < Len; ++i )
        if( i < FFT_Index[ i ] )
            swap( A[ i ], A[ FFT_Index[ i ] ] );
    for( long long Half = 1; Half < Len; Half <<= 1 ) 
        for( long long i = 0; i < Len; i += Half << 1 ) 
            for( long long j = 0; j < Half; ++j ) 
                MyComplex Temp = Omega[ ( Len / Half / 2 ) * j ] * A[ i + j + Half ];
                MyComplex T = A[ i + j ];
                A[ i + j ] = Temp + T;
                A[ i + j + Half ] = T - Temp;
            
    return;

long long FFT_MAIN() 
    FFT( FFT_A, UpTo ); FFT( FFT_B, UpTo );
    for( long long i = 0; i < UpTo; ++i ) FFT_A[ i ] = FFT_A[ i ] * FFT_B[ i ];
    for( long long i = 0; i < UpTo; ++i ) Omega[ i ].Imagine = -Omega[ i ].Imagine;
    FFT( FFT_A, UpTo );
    for( long long i = 0; i < UpTo; ++i ) FFT_A[ i ].Real = floor( FFT_A[ i ].Real / UpTo + 0.5 );
    return UpTo;

void Init( long long Need)    //FFT初始化
    memset( FFT_Index, 0, sizeof( FFT_Index ) );
    memset( Omega, 0, sizeof( Omega ) );
    long long Len = 2 * Need - 1;
    for( UpTo = 1; UpTo <= Len; UpTo <<= 1 );
    for( long long i = 0; i < UpTo; ++i ) 
        FFT_Index[ i ] = ( FFT_Index[ i >> 1 ] >> 1 ) | ( ( i & 1 ) * ( UpTo >> 1 ) );
    for( long long i = 0; i < UpTo; ++i )
        Omega[ i ] = ( MyComplex )  cos( 2.0 * Pi * i / UpTo ), sin( 2.0 * Pi * i / UpTo ) ;
    return;

void Work() 
    scanf( "%lld", &n );
    for( long long i = 1; i <= n; ++i ) scanf( "%lld", &A[ i ] );
    m = 0; for( long long i = 1; i <= n; ++i ) m = max( m, A[ i ] );
    Init( m + 1 );

    memset( FFT_A, 0, sizeof( FFT_A ) );
    for( long long i = 1; i <= n; ++i ) FFT_A[ A[ i ] ].Real += 1;
    memcpy( FFT_B, FFT_A, sizeof( FFT_B ) );

    FFT_MAIN();   //FFT求两条边的长度和的所有情况

    for( long long i = 1; i <= n; ++i ) FFT_A[ A[ i ] + A[ i ] ].Real -= 1.0;//两条相同的边是不被允许的
    for( long long i = 0; i < UpTo; ++i ) FFT_A[ i ].Real /= 2.0;//不同顺序选两条边认为是相同情况

    memset( Sum, 0, sizeof( Sum ) );
    for( long long i = 1; i < UpTo; ++i ) Sum[ i ] = Sum[ i - 1 ] + FFT_A[ i ].Real;//前缀和预处理
    long long Ans = 0;
    sort( A + 1, A + n + 1 );
    for( int i = 1; i <= n; ++i ) 
        long long Temp = Sum[ UpTo - 1 ] - Sum[ A[ i ] ];//和大于当前枚举边的所有情况
        Temp -= n - 1;//枚举的这条边不能再被选
        Temp -= ( n - i ) * ( i - 1 );//减去其中一条边大于当前边的情况
        Temp -= ( n - i - 1 ) * ( n - i ) / 2;//减去两条边都大于当前边的情况
        Ans += Temp;
    
    long long Total = 1LL * n *( n - 1 ) * ( n - 2 ) / 6;
    printf( "%.7lf\n", 1.0 * Ans / Total );
    return;