A partition of a positive integer n is an non-increasing array of positive integers
a[1] , a[2] , ... , a[m]
satisfying
a[1] + a[2] + ... +a[m] = n.
m is called the length of this partition.
We can list all the partitions of n in an specified order. For example, if we use the rule by which a lexicography sorts all the English words, it's called the lexicographic order. Another way, if we use the rule by which the C language compares strings, it's called the reverse lexicography order. And there is also a colex order.
To generate all the partitions of a integer n, we have a nice algorithm proposed by Stojmenovic, which has already been included in Knuth's book.
To generate all the partitions of n with length exactly m, we can use colex order,this algorithm is also included in Knuth's book.
To generate all the partitions of n with all their elements not exceeding k, we can use the algorithm in 1, just changing its initial condition and exit condition of the cycle.
Here comes my question: how to generate those partitions that have their length exactly be m, and their elements not exceeding k?
Here m and k are constants . Of course, a partition with its elements not exceeding k is equivalent to its first element not exceeding k.
Oh, I think I have solved it. for
a[1] + a[2] + ... + a[m] = n
can be written as
(k+1-a[1]) + (k+1-a[2]) + ... + (k+1-a[m]) = m(k+1)-n
and the latter one is just a reversed partition of m(k+1)-n !