vijos-p1077 Clone Dragon (Find pattern + exponential parent function + python)

P1077Clone Dragon Accepted Tags: [show tags]<textarea id="code" class="textbox" style=""></textarea>Describe

Now the dragon cloning has become possible, the Dragon gene is composed of ACTG letters, and the Dragon's genes have the following characteristics:
1, a in the gene appears for an even number of times (including 0);
2, the same is true of C;
There are 6 conditions to meet when n=2:
Tt,tg,gt,gg,aa,cc
You just give the last two digits of the number of genes that meet the criteria;

Format input Format

The input file gives a number of n (1<=n<=10^9). End with 0.

Output format

For the input n, the last two digits of the number of strings that satisfy the condition!

Example 1 sample input 1[copy]

1230

Sample output 1[Copy]

2620

Source

Huyichen

Find the pattern, the exponential parent function will be uploaded later

The 0 here should be treated with special treatment, compared to pits

Python is just a recursive type.

#!/usr/bin/env python3#-*-coding:utf-8-*-import mathimport sysfor cin in Sys.stdin:    n = Long (CIN)    if not n:bre AK    N-= 1    a = Math.ceil (POW (2, N, 0) * (POW (2, N, 00) + 1))    a = Long (a)    if a% = =:        print ' '    Else:        print a% 100

C++:

#include <cstdio> #include <cstring> #include <algorithm>using namespace Std;typedef long Long ll;int A [one] = {2,6,20,72,72,56,60,12,92,56};int B[21] = {0,52,12,56,40,92,32,56,80,32,52,56,20,72,72,56,60,12,92,56};int Main () {    int n;    while (~ scanf ("%i64d", &n), n) {        if (n <=) {            n--;            printf ("%d\n", A[n]);        } else {            if (b[(n-11)%] = = 0) printf ("00\n");            else printf ("%d\n", b[(n-11));        }    }    return 0;}

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vijos-p1077 Clone Dragon (Find pattern + exponential parent function + python)