One of the key counts (recursive) hdu1438

One of the key countsTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 1471 Accepted Submission (s): 622


Problem description A key has n slots, Groove deep 1,2,3,4. Each key has at least 3 different depths and at least 1 pairs of connected slots have a depth difference of 3. To find the total number of such keys.
Input not in the subject
Output to n>=2 and n<=31, outputs the total number of keys that meet the requirements.
Sample Output

N=2:0N=3:8N=4:64N=5:360..............N=31: ... Note: Adapted according to PKU Judge Online 1351 number of locks or Xi ' an 2002, where n<=16

Reprint please specify the Source: Search & Star Children

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1438

/*1: If x is the key, then X1/2/3/4 is also, so a[i]=a[i-1]*42: If x is not, X2/3 is the x is composed of 1/4, but to remove all 1 and all 4, so a[i]+= (2^ (I-1) 2) * * Suffix 2 or 3 plus it becomes the key, Must be not satisfied need 3 different depth groove this one, so x must be composed of 1/4 3 if X is not X1/4 is. Then X=y (m=x1/4=y) (4/1) (1/4) The front I-2 bit can be 1234, but to remove all the cases of 1/4 is the case that X is the key and X is the end of 1/4. We use the B[i] array to represent the number of I bits ending at 1/4   temp= (4^ (i-2) -2^ (i-2)) * 2–b[i-1]; b[i]=a[i-1]*2+temp;*/#include <stdio.h># Define LL long Long#include<string.h>ll a[35],b[35]; ll Ppow (ll n,ll m) {    ll c=1;    while (m)    {        if (m&1) c=c*n;        N=n*n;        m>>=1;    }    return c;} int main () {    a[0]=a[1]=a[2]=0;    b[0]=b[1]=b[2]=0;    LL temp=0;    printf ("n=2:0\n");    for (int i=3;i<=31;i++)    {        temp= (Ppow (4,i-2)-ppow (2,i-2)) *2-b[i-1];//       printf ("temp=%lld\t", temp );        B[i]=a[i-1]*2+temp;        a[i]=a[i-1]*4+ (Ppow (2,i-1)-2) *2+temp;        printf ("n=%d:%lld\n", I,a[i]);    }    return 0;}

One of the key counts (recursive) hdu1438