Fibonacci Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15238 Accepted Submission(s): 7140
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.
Print the word "no" if not.
Sample Input
0 1 2 3 4 5
Sample Output
no no yes no no no
Print the word"yes" if 3 divide evenly into F(n);Print the word"no" if not.
这里m取值为3,则可将公式条件演变为:
综上所述,可得到以下对应关系:F(0)= 1, F(1) = 2, F(n) = ( F(n-1) + F(n-2) )( mod 3) (n>=2).
index 0 1 2 3 4 5 6 7 8 9 10 11 12 13
value 1 2 0 2 2 1 0 1 1 2 0 2 2 1
print no no yes no no no yes no no no yes no no no
这样我们就得到了如下规律:从第2个开始每隔4个循环一次。
#include<stdio.h>
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
if((n-2)%4!=0)
printf("no\n");
else
printf("yes\n");
}
return 0;
}