Simplify hashfull calculation.

We can simplify the calculation of the hashfull info by looping over exact 1,000 entries,
and then divide the result by ClusterSize. Somewhat memory accesses, somewhat more accurate.

Passed non-regression LTC
https://tests.stockfishchess.org/tests/view/5e30079dab2d69d58394fd5d
LLR: 2.94 (-2.94,2.94) {-1.50,0.50}
Total: 30125 W: 3987 L: 3926 D: 22212
Ptnml(0-2): 177, 2504, 9558, 2642, 141

closes #2523

No functional change.

src/tt.cpp

@@ -148,9 +148,9 @@ TTEntry* TranspositionTable::probe(const Key key, bool& found) const {
 int TranspositionTable::hashfull() const {
 
   int cnt = 0;
-  for (int i = 0; i < 1000 / ClusterSize; ++i)
+  for (int i = 0; i < 1000; ++i)
       for (int j = 0; j < ClusterSize; ++j)
           cnt += (table[i].entry[j].genBound8 & 0xF8) == generation8;
 
-  return cnt * 1000 / (ClusterSize * (1000 / ClusterSize));
+  return cnt / ClusterSize;
 }