Nth number of ugly

I think for two days have no idea ah ah ah ah oh I was so stupid. Read his tag, said the use of dynamic planning, so deliberately read the introduction of the algorithm in the dynamic planning part. However, the first step is to construct a reasonable data structure, and the second is to solve it in the form of recursion. Visible water impermanence, dynamic planning is not a simple formula can be solved. So how do you deal with it?

The first is that when dynamic planning is used, it is the case that the results from the previous results are then stored after the results have been solved. The former implies its recursive structure, which has the meaning of space-changing time.

Clear this goes into the second question, what is the ugly number?

In addition to 1, the number consisting of only 2 3 53 vegetarian Factors is the number of ugly. Since the vegetarian factor can only have 2 3 5, then the number of ugly in the back must be obtained by multiplying the number of 2/3/5 in front.

The next number is calculated from the front, and the previous ugly number is calculated and stored in the array, which is the two points mentioned earlier recursion and space change time.

So how is the new ugly number likely to be formed? Three kinds: 2 *A/3 * B/5 * C Therefore, three markers are set, indicating the position of a, B, and C respectively.

Among them, the smallest number becomes the new ugly number. And after it is selected as the new ugly number, the next new ugly number must be larger than this number (nonsense) so the corresponding tag is shifted back to ensure that the next new ugly number of three choices are greater than the current maximum number of ugly.

With the above instructions, you can understand the algorithm very well:

intNthuglynumber (intN) {if(N <1){        return 0; } Vector<int> ugliness{1}; Auto Index2=0; Auto Index3=0; Auto Index5=0; Auto Savenewuglyincreaseit= [&] {AutoConstVal2 = ugliness[index2] *2; AutoConstVAL3 = ugliness[index3] *3; AutoConstVAL5 = ugliness[index5] *5; AutoConstnewugly =min (val2, min (Val3, VAL5)); if(newugly = =val2) {            ++Index2; }        if(newugly = =Val3) {            ++index3; }        if(newugly = =VAL5) {            ++index5;    } ugliness.push_back (newugly);        };  for(inti =1; I! = N; ++i)    {Savenewuglyincreaseit (); }    returnugliness.back ();}

Nth number of ugly