Fibonacci Again
链接:http://acm.hdu.edu.cn/showproblem.php?pid=1021
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
0
1
2
3
4
5
Sample Output
no
no
yes
no
no
no
该题只需直接计算并对三取模即可
#include <bits/stdc++.h>
using namespace std;
int f[1000005];
int main()
{
f[0]=1,f[1]=2;
for(int i=2;i<1000000;i++){
f[i]=(f[i-1]+f[i-2])%3;
}
int n;
while(cin>>n){
if(f[n]!=0)cout<<"no"<<endl;
else cout<<"yes"<<endl;
}
return 0;
}