ACM1021:Fibonacci Again
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
-------------------------------------------------------------------------------
#include <stdio.h>
#include <string.h>
#define N 1000001
int ans[N];
int main()
{
int i;
int n;
memset(ans, 0, sizeof(ans));
ans[0] = 7 % 3;
ans[1] = 11 % 3;
for (i = 2; i <= N; i++)
ans[i] = (ans[i - 1] + ans[i - 2]) % 3;
while (scanf("%d", &n) != EOF)
{
if (ans[n] == 0)
printf("yes\n");
else
printf("no\n");
}
return 0;
}

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