HDU 2045 Not Easy Series (3)--lele RPG puzzle

Ideas:

1. If the color of the former n-1 is eligible, then there are only 1 ways to join one because of the difference in the first and last, s[n] = s[n-1]

2. If a string consisting of a former n-1 is not met, then it is lawful to join one. That is, because the end of the same caused by the illegal, then the string of the former n-2 is bound to be legal. At this point there are 2 ways to add the nth bit. ie s[n] = 2*s[n-2]

So the AC code:

<span style= "FONT-FAMILY:KAITI_GB2312;FONT-SIZE:24PX;" > #include <stdio.h>int main () {    int n,i;    __int64 a[55];    a[1]=3;    a[2]=6;    a[3]=6;    for (i=4;i<=50;i++)        a[i]=a[i-1]+2*a[i-2];    while (scanf ("%d", &n)!=eof)    {        printf ("%i64d\n", A[n]);    }    return 0;} </span>

HDU 2045 Not Easy Series (3)--lele RPG puzzle