I can't really make R < B all the time.
In my solution, dp[a][b][r] means the maximum set of pair(x, y) such that
sum(x) <= b
sum(y) <= r
and all x >= a
so
dp[a][b][r] = max {dp[a + 1][b - (j + 1) * a][r - j * (j + 1) / 2] + j + 1}
assuming that the set contains (a, 0) (a, 1) ..... (a, j)
which means, dp[a][b][r] is different from dp[a][r][b]
The answer is simply dp[0][b][r] given b and r
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