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Fibonacci Again

2024-10-30 21:56:39   206人阅读

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39792 Accepted Submission(s): 19088


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
源代码一：
#include <stdio.h>#include <stdlib.h>int f[1000000];void fac(){   int i;   f[0]=7%3,f[1]=11%3;   for(i=2;i<1000000;i++)     f[i]=(f[i-1]%3+f[i-2]%3)%3;                  }int main(){  int n;  fac();  while(scanf("%d",&n)!=EOF)  {    if(n<2)      printf("no\n");    else    {     if(f[n]==0)       printf("yes\n");     else      printf("no\n");        }                            }  system("pause");  return 0;    }
源代码二：
#include <stdio.h>#include <stdlib.h>int main(){    int n;  int fac[8]={1,2,0,2,2,1,0,1};  while(scanf("%d",&n) != EOF)  {    if(!fac[n%8])       printf("yes\n");    else       printf("no\n");  }  system("pause");  return 0;}

Fibonacci Again


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