UVa 1647 (Recursive) Computer transformation

Test instructions

There is a 01 string, each step will be all 0 to 10, will all 1 into 01, the string is the first 1.

After the nth step, the number of 00

Analysis:

At the beginning of the time is still relatively chaotic, I also tangled up 000 is a 00 or 2 00

After you finally want to understand, there will not be such a substring.

Summed up several points:

• After the nth step, the string length is 2n, and the number of 0 and 1 is equal, respectively 2n-1
• 1 after a two-step change of 1001, so each 1 after a two-step change will get a 00, and this 00 is left and right on both sides of a 1-pack, will not be with other numbers to gather extra 00
• 0 after a two-step change of 0110, so 00 will become011 110, so 00 changes two times still have 00

Finally, the conclusion is drawn:

so that f (n) is the number of 00 in the string after n changes, there is a recursive relationship f (n+2) = f (n) (two times the number of 00 before the change) + 2n-1 (two changes before the number of 1)

Because n can be as big as 1000, so use high precision.

1#include <iostream>2#include <cstdio>3 using namespacestd;4 5 inta[1005][ Max], b[1005][ Max];6 7 intMain ()8 {9a[0][0] = a[1][0] =1;Ten      for(inti =2; I <= +; i++) One          for(intj =0; J <135; J + +) A         { -A[I][J] + = a[i-1][J] + a[i-1][j]; -B[I][J] + = b[i-2][J] + a[i-2][j]; thea[i][j+1] + = A[i][j]/10000; A[I][J]%=10000; -b[i][j+1] + = B[i][j]/10000; B[I][J]%=10000; -         } -  +     intN; -      while(SCANF ("%d", &n) = =1) +     { A         inti; at          for(i =135; i >0&& B[n][i] = =0; i--); -printf"%d", B[n][i]); -          for(i--; I >=0; i--) printf ("%04d", B[n][i]); -printf"\ n"); -     } -  in     return 0; -}

code June

UVa 1647 (Recursive) Computer transformation