Created
May 24, 2011 17:42
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BitTorrent task solution
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#include<stdio.h> | |
const int BOWLS = 40; | |
const int ORANGES = 9; | |
int count = 0; | |
// All the places in which an orange can't be placed are 1's | |
int used[BOWLS]; | |
// A list of the oranges already placed. | |
int oranges[ORANGES]; | |
int turn = 0; | |
int y; | |
int tmp_index; | |
template <int turn> | |
inline void check(int x) { | |
if (turn == ORANGES) { | |
++count; | |
return; | |
} | |
for(; x < BOWLS - (ORANGES - turn - 1); ++x) { | |
// Only check available positions. | |
if (!used[x]) { | |
// Mark this position as used. | |
used[x] += 1; | |
// This position is called "B" | |
// For every A, make sure no C exists such that |A-B| = |B-C| | |
// I.e. mark them as used even though no oranges are placed there. | |
for(y = 0; y < turn; ++y) { | |
tmp_index = x + x - oranges[y]; | |
if (tmp_index >= 0 && tmp_index < BOWLS) { | |
used[tmp_index] += 1; | |
} | |
} | |
// The orange on this turn is placed on "x" | |
oranges[turn] = x; | |
check<turn+1>(x+1); | |
// Revert the "used" markers placed this turn. | |
for(y = 0; y < turn; ++y) { | |
tmp_index = x + x - oranges[y]; | |
if (tmp_index >= 0 && tmp_index < BOWLS) { | |
used[tmp_index] -= 1; | |
} | |
} | |
used[x] -= 1; | |
} | |
} | |
} | |
template <> | |
inline void check<ORANGES>(int x) { | |
++count; | |
} | |
int main() { | |
check<0>(0); | |
printf("%d\n", count); | |
} |
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