【牛客多校第十场】Coffee Chicken

题目描述

Dr. JYY has just created the Coffee Chicken strings, denoted as S(n). They are quite similar to the Fibonacci soup --- today's soup is made by mingling yesterday's soup and the day before yesterday's soup:
- S(1)="COFFEE"S(1) = \texttt{"COFFEE"}S(1)="COFFEE";
- S(2)="CHICKEN"S(2) = \texttt{"CHICKEN"}S(2)="CHICKEN";
- S(n) = S(n-2) :: S(n-1), where :: denotes string concatenation.
The length of S(n) quickly gets out of control as n increases, and there would be no enough memory in JYY's game console to store the entire string. He wants you to find 10 contiguous characters in S(n), starting from the kth character.

描述输出

The first line of input is a single integer T (1≤T≤1000)(1 \leq T \leq 1000)(1≤T≤1000), denoting the number of test cases. Each of the following T lines is a test case, which contains two integers n, k (1≤n≤500,1≤k≤min⁡{∣S(n)∣,1012})(1 \leq n \leq 500, 1 \leq k \leq \min\{|S(n)|, 10^{12}\})(1≤n≤500,1≤k≤min{∣S(n)∣,1012}). |S| denotes the length of string S.

输出描述

For each test case, print the answer in one line. If there are no enough characters from the kth character, output all characters until the end of the string.

输入

2
3 4
2 7

输出

FEECHICKEN
N

题目思路：

可以用递归模拟，因为这是一个类似斐波那契额数列，当前这个位置是由上一次前一次推出来的，所以当前位置k就等于a[n-2]的那个位置或者是k=k-a[n-1]；k>a[n-2]长度,那么就相当于求 a[n-1]的第k-a[n-2]个数,如果k<=a[n-2]那就等于求a[n-2]的第k个数，感谢大佬溺生的思路，不太懂的去看看！

AC代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+5;
ll a[700];
void make()
{a[1]=6;a[2]=7;for(int i=3;i<=60;i++){a[i]=a[i-1]+a[i-2];}
}
int main()
{make();
//  printf("%lld\n",a[60]);char x[2][30]={"COFFEE","CHICKEN"};int t;scanf("%d",&t);while(t--){ll n,k;scanf("%lld%lld",&n,&k);if(n>=60){if(n%2)   n=59;else n=58;}for(ll i=k;i<=a[n]&&i<k+10;i++){ll w=i,v=n;while(v!=1&&v!=2){if(w>a[v-2]){w=w-a[v-2];v-=1;}elsev-=2;}if(v==1)printf("%c",x[0][w-1]);else if(v==2)printf("%c",x[1][w-1]);}printf("\n");}return 0;
}

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