UVA, 580 Critical Mass

Test instructions: Give you u and l string 3 consecutive U for dangerous goods, give you x length of string, to find the number of dangerous

Idea: Recursion

Number of security solutions: m[i]=m[i-1]+m[i-2]+m[i-3] Fibonacci sequence Upgrade (╯‵-′) ╯︵┻━┻

1#include <iostream>2#include <cstdio>3 using namespacestd;4 intN;5 #defineMAXN 516 Long LongM[MAXN];7 BOOLdatecin ()8 {9     if(SCANF ("%d", &n)! =EOF)Ten     { One         if(n==0) A             return false; -         return true; - }ac the     return false; - } -  - voiddatecal () + { -  +m[0]=1, m[1]=1; A      for(intI=2; i<maxn;i++) atm[i]=m[i-1]+m[i-2]; - } -  - voidshowres () - { -printf"%lld\n", M[n]); in } - intMain () to { + datecal (); -      while(Datecin ()) the     { * showres (); $     }Panax Notoginseng     return 0; -}

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UVA, 580 Critical Mass