(Hdu step 3.1.6) statistical problem (number of solutions that continuously walk left and right and go up n steps), hdu3.1.6

(Hdu step 3.1.6) statistical problem (number of solutions that continuously walk left and right and go up n steps), hdu3.1.6

Make an advertisement for yourself before writing a question ~.. Sorry, I hope you can support my CSDN video courses at the following address:

Http://edu.csdn.net/course/detail/209

Question:

Statistical problems

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission (s): 732 Accepted Submission (s): 466

Problem Description in an infinitely large two-dimensional plane, we make the following assumptions: 1. Only one cell can be moved at a time; 2. You cannot go backward (If your destination is "up", you can go left, right, or up, but not down ); 3. The passing grid collapsed immediately and could not go for the second time; Calculate the number of different solutions in n steps (two steps are considered different if one step is different ).

Input first gives a positive integer C, indicating that there are group C Test Data In row C, each row contains an integer n (n <= 20), which indicates that n steps are required.

For the Output, Please program the total number of different solutions that take n steps; The output of each group occupies one row.

Sample Input 212

Sample Output 37

Authoryifenfei

Source Shaoxing top school of Information Technology-Program Design Competition of the second Computer Culture Festival

Recommendyifenfei

Question Analysis:

Number of n-grams f (n) = 2 * f (n-1) + f (n-2 ). How did this happen. Use a (n) to represent the number of solutions in step n and B (n) to represent

Number of solutions taken around step n. So a (n) = a (n-1) + B (n-1) (here, Why can't B (n-1) be multiplied by 2? Because the left-to-right direction cannot be used at the same time ). B (n) = 2 * a (n-1) + B (n-1) (a (n-1) * 2 is because, in step n, you can choose to go left or right. These two solutions are also available. For B (n-1), because if you go in the direction of step n-1, you can only go in the direction of step n)

The Code is as follows:

/** F. cpp ** Created on: February 5, 2015 * Author: Administrator */# include <iostream> # include <cstdio> using namespace std; const int maxn = 22; long dp [maxn]; void prepare () {dp [1] = 3; dp [2] = 7; int I; for (I = 3; I <maxn; ++ I) {dp [I] = 2 * dp [I-1] + dp [I-2] ;}} int main () {prepare (); int t; scanf ("% d ", & t); while (t --) {int n; scanf ("% d", & n); printf ("% lld \ n", dp [n]);} return 0 ;}