Fibonacci Again

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

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

一个斐波那契数列数列，且F[0]=7，F[1]=11，题目问F
能否被3整除，能整除输出yes，否则输出no。有公式(a+b)%p=(a%p+b%p)%p，根据这个公式，就可以算出整个数列上每一项MOD3，如果这一项为0，输出yes，否则输出no。

C

#include <stdio.h>
#pragma warning(disable:4996)
int f[1000000]={1,2};
int main(void)
{
int i=2,n;
while(scanf("%d",&n)!=EOF)
{
for(;i<=n;i++)           //i=2的初始化放在声明那里，可以节省一点时间，数组只会遍历一遍
f[i]=(f[i-1]%3+f[i-2]%3)%3;
if(f[n]==0)
printf("yes\n");
else
printf("no\n");
}
return 0;
}

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