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 = ;
const LL MOD = (1e9) + ;

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 = , y = ; }
    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 = ;
        for ( int i = ; 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 = , j = ; i < n ; i += , ++j )
            a[i] = b[j];
        for ( i = , j = n - ; i < n; i += , --j )
            a[i] = b[j];

//        for ( int i = 0; i < n; ++i )
//            printf( "%I64d ", a[i] );
//        puts("");

        LL p = ;
        for ( int i = ; i < n; ++i )
        {
            LL tmp = ( all * GetInverse( ( a[i] * a[i - ] ) % MOD ) ) % MOD;
            p += ( min( a[i], a[i - ] ) * tmp ) % MOD ;
            p %= MOD;
        }
        all *= n;
        all %= MOD;
//        printf("p = %I64d\n", p );
        LL ans = ( all - p + MOD ) % MOD;
        printf( "%I64d\n", ans );
    }
    return ;
}