HDU 4655 Cut Pieces 找规律+简单计数

解法参考：http://blog.csdn.net/a601025382s/article/details/9840125

 

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <algorithm>



using namespace std;



#define LL long long int



const int MAXN = 1000010;

const LL MOD = (1e9) + 7;



int n;

LL a[MAXN];

LL b[MAXN];



void ExGcd( LL a, LL b, LL& d, LL& x, LL& y )

{

    if ( !b ) { d = a, x = 1, y = 0; }

    else

    {

        ExGcd( b, a % b, d, y, x );

        y -= x * ( a / b );

    }

}



LL GetInverse( LL num )

{

    LL d, x, y;

    ExGcd( num, MOD, d, x, y );

    return ( x % MOD + MOD ) % MOD;

}



int main()

{

//    freopen( "1001.in", "r", stdin );

//    freopen( "s.out", "w", stdout );

    int T;

    scanf( "%d", &T );

    while ( T-- )

    {

        scanf( "%d", &n );

        LL all = 1;

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

        {

            scanf( "%I64d", &a[i] );

            b[i] = a[i];

            all *= a[i];

            all %= MOD;

        }



        //printf( "all = %I64d\n", all );



        sort( b, b + n );

        int i, j;

        for ( i = 0, j = 0; i < n ; i += 2, ++j )

            a[i] = b[j];

        for ( i = 1, j = n - 1; i < n; i += 2, --j )

            a[i] = b[j];



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

//            printf( "%I64d ", a[i] );

//        puts("");



        LL p = 0;

        for ( int i = 1; i < n; ++i )

        {

            LL tmp = ( all * GetInverse( ( a[i] * a[i - 1] ) % MOD ) ) % MOD;

            p += ( min( a[i], a[i - 1] ) * tmp ) % MOD ;

            p %= MOD;

        }

        all *= n;

        all %= MOD;

//        printf("p = %I64d\n", p );

        LL ans = ( all - p + MOD ) % MOD;

        printf( "%I64d\n", ans );

    }

    return 0;

}