【递推】Fibonacci Again
作者：互联网
描述
There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).
输入
Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).
输出
Print the word “yes” if 3 divide evenly into F(n).
Print the word “no” if not.
样例输入
0
1
2
3
4
5
样例输出
no
no
yes
no
no
no
题目来源
HDOJ
分析：斐波那契递推。
代码：
#include<bits/stdc++.h>
using namespace std;
int f[1000001];
void Fibo()
{
f[0]=7;f[1]=11;
for (int i=2;i<1000001;i++)
f[i]=(f[i-1]+f[i-2])%3;
}
int main()
{
Fibo();
int n;
while(cin>>n)
{
if (f[n]%3==0)
cout<<“yes”<<endl;
else
cout<<“no”<<endl;
}
return 0;
}
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标签：Again,no,int,样例,Fibo,Fibonacci,Print,yes,递推 来源： https://blog.csdn.net/Skynamer/article/details/104079423