More than 2019 cattle off summer school camp (tenth field) Coffee Chicken-- recursive

The meaning of problems

$ S (1) = "COFFEE", S (2) = "CHICKEN" $, $ S (n) = S (n-2) + S (n-1) $, please output $ S (n) $ in starting from the $ k $ characters 10 characters (if 10 output is insufficient to the end of can), $ 1 \ leq n \ leq 500, 1 \ leq k \ leq min ({| S (n) |, 10 ^ {12}}) $.

analysis

Note $ f (i) $ represents $ S (i) $ length, $ f (i) = f (i-2) + f (i-1) $.

Recursive call $ solve (n, k) $ represents $ S (n) $ of $ k $ characters.

If $ n \ leq 2 $ return directly answer,

$ N> 2 $, if $ k> f (i-2) $, calls $ solve (n-1, kf (n-2)) $; otherwise the call $ solve (n-2, k) $.

Note that Japanese sentence $ k> f (n) $.

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn = 510;
char s1[10] = "COFFEE";
char s2[10] = "CHICKEN";
ll n, k;
ll f[maxn];

void init()
{
    f[1] = 6;f[2] = 7;
    int i;
    for(i = 3; I <MAXN; I ++ ) 
    { 
         F [I] = F [I- 2 ] + F [I- . 1 ];
          IF (F [I]> 1e12 + 10 ) BREAK ; 
    } 
    for ( int J = I + . 1 ; J < MAXN; J ++)   // more than 1e12, directly by the equal process, or will overflow 
        F [J] = F [I]; 
} 

char Solve (LL n-, LL K) 
{ 
    IF (K> F [n-])   return  ' A ' ;
     IF (n-== . 1 )   return S1 [- K- . 1 ];
     IF(n == 2)  return s2[k-1];

    if(k > f[n-2])  return solve(n-1, k-f[n-2]);
    else  return solve(n-2, k);
}

int main()
{
    init();
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%lld%lld", &n, &k);
        for(int i = 0;i < 10;i++)
        {
            char ch = solve(n, k+i);
            if(ch == 'A')  break;
            printf("%c", ch);
        }
        printf("\n");
    }
}