Fibonacci Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44439 Accepted Submission(s): 21214
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
解题思路:实在没啥说的,就是找寻环节,直接上代码吧,其实还可以用矩阵乘法做,感觉有点麻烦,但是为了学会矩阵乘法,我觉定要用矩阵做一下,当然这是待会的事情了,嘿嘿:
/*
2015 - 8 - 13
Author: ITAK
今日的我要超越昨日的我,明日的我要胜过今日的我,
以创作出更好的代码为目标,不断地超越自己。
*/
#include <iostream>
#include <cstdio>
using namespace std;
int data[30];
int main()
{
/*
打表看一下,找寻环节
data[0] = 7;
data[1] = 11;
for(int i=2; i<30; i++)
data[i] = data[i-1]+data[i-2];
for(int i=0; i<30; i++)
if(data[i]%3 == 0)
cout<<i<<endl;
*/
int m;
while(~scanf("%d",&m))
{
if(m%4 == 2)
puts("yes");
else
puts("no");
}
return 0;
}
时间: 2024-08-29 12:10:34