2019 cattle off more school tenth field B.Coffee Chicken (recursive)

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The meaning of problems

The first string of "COFFEE", the second string is "CHICKEN", the n-th string from its front 2 and a front connected to each other (i.e., s [n] = s [n -2] + s [n-1], note that s [n-2] in front of s [n-1]), the query Q from the n-th k-th character string of 10 consecutive characters beginning what (if not there are 10 characters but reached the end of the string to stop)

$1$

 answer

Code

#include <bits / STDC ++ H.>
 the using  namespace STD; 
typedef Long  Long LL;
 const LL + 1e12 inf = 10 ; // K maximum 1e12, coupled to the output 10, the value may inf + 10 1e12 
const  int MAXN 5e2 + = 10 ; 
LL S [MAXN]; 
String ANS1 = " COFFEE " , ANS2 = " CHICKEN " ;
 char Solve (n-LL, LL K) 
{ 
    IF (n-== . 1 ) return ANS1 [- K- . 1 ]; // array starts from 0, 1 and k from the beginning, so the need -1 
    the else  IF(n-== 2 ) return ANS2 [- K- . 1 ];
     IF (K> S [N- 2 ])
         return Solve (N- . 1 , KS [N- 2 ]);
     the else 
        return Solve (N- 2 , K) ; 
} 
int main () 
{ 
    / * pretreatment string length * / 
    S [ . 1 ] = . 6 , S [ 2 ] = . 7 ;
     for ( int I = . 3 ; I <= 500 ; I ++ ) 
        S [I] = S [I- 2] + S [I- . 1 ] <? S INF = [I- 2 ] + S [I- . 1 ]: INF; // if s [i]> 1e12 + 10, the rear have made equal to 1e12 + 10, or overflow 
        
    int T; 
    CIN >> T; 
    LL n-, K; 
    the while (T-- ) 
    { 
        CIN >> >> n- K;
         for (I = K LL; I <K + 10 && I <= S [n-]; I ++ ) // a outputs, to the 10 or greater than the total length of the string is stopped (Note that for i longlongs) 
            COUT << Solve (n-, i); 
        COUT << endl; 
    } 
    return  0 ; 
}