杭电 HDU 1021 Fibonacci Again

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41156    Accepted Submission(s): 19705

Problem Description

There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).

 

Input

Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).

 

Output

Print the word "yes" if 3 divide evenly into F(n).

Print the word "no" if not.

 

Sample Input

012345

 

Sample Output

nonoyesnonono

 

Author

Leojay

 

 

我说这个题目想当有意思，首先上去就做，压根就没考虑数组溢出n到几百就不行了；最关键的一点是题目要求一种是与否的判断，那就好好找找规律，结果直接让他输出100个f【n】看是否能被3整除的数据是如何分布的，立刻就遇到了惊喜！请看：

#include<iostream>using namespace std;int main(){int n;int *ls=new int [10000000];ls[0]=7;ls[1]=11;for(int i=2;i<100;i++){ls[i]=ls[i-1]+ls[i-2];/*if(ls[i]%3==0)      cout<<"yes"<<" ";  else  cout<<"no"<<" ";*/}while(cin>>n){printf((n-1)%4==1?"yes\n":"no\n");}//printf(ls[n]%3?"no":"yes");return 0;}

结果AC代码能猜出几行么！！呵呵呵

#include<iostream>using namespace std;int main(){int n;while(cin>>n){printf((n-1)%4==1?"yes\n":"no\n");}return 0;}