HDU 4655 Cut Pieces 找规律+简单计数

#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;
}