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 ;
}
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