Fibonacci Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 61009 Accepted Submission(s): 28508
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
//题目要求根据公式判断 fn 是否能够被3整除，可以的话输出 yes 不可以的话输出 no
//但是当出现测试数据达到百万，肯定是找规律的题目，耐下心来仔细研究一下就会发现 n % 4 == 2 的时候都可以被3整除
#include <iostream>
using namespace std;
int main()
{
int n;
while(scanf("%d",&n) != EOF)
{
if(n % 4 ==2)
cout << "yes" << endl;
else
cout << "no" << endl;
}
}//或者利用公式 枚举也可以的
#include <iostream>
#include <cstdio>
using namespace std;
const int MAX_N = 1000010;
int f[MAX_N];
void init(){
f[0] = 7; f[1] = 11;
for(int i = 2 ; i < MAX_N ; i ++){
f[i] = ( f[i-1]%3 + f[i-2]%3 )%3;
}
}
int main(){
init();
int n;
while(~scanf("%d",&n)){
if( f[n] == 0 ) printf("yes\n");
else printf("no\n");
}
return 0;
}
