https://vjudge.net/contest/67836#problem/A
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
时间复杂度：$O(10^6)$
代码：
#include <bits/stdc++.h>
using namespace std; const int maxn = 1e6 + 10;
int Fib[maxn]; int main() {
Fib[0] = 7, Fib[1] = 11;
for(int i = 2; i < maxn; i ++)
Fib[i] = (Fib[i - 1] + Fib[i - 2]) % 3; //同余定理
int n;
while(~scanf("%d", &n)) {
if(Fib[n] % 3)
printf("no\n");
else
printf("yes\n");
}
return 0;
}