Fibonacci Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43539 Accepted Submission(s): 20797
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
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
const int Max=1100000;
int Arr[Max];
int main()
{
memset(Arr,0,sizeof(Arr));
Arr[0]=1;
Arr[1]=2;
for(int i=2;i<Max;i++)
{
Arr[i]=(Arr[i-1]+Arr[i-2])%3;
}
int n;
while(~scanf("%d",&n))
{
if(!Arr[n])
{
printf("yes\n");
}
else
{
printf("no\n");
}
}
return 0;
}
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