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Fibonacci Again

时间:2015-01-10 12:42:45      阅读:161      评论:0      收藏:0      [点我收藏+]

标签:printf   数据   测试   

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

标签:printf   数据   测试   

原文地址:http://blog.csdn.net/zchlww/article/details/42581121

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