Fibonacci Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58267 Accepted Submission(s): 27275
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
Author
Leojay
分析：有两种方法：第一种是找规律，除4余2的就输出yes。否则no.这种方法确实很神奇，想不到没关系！先给出第一种方法的ＡＣ代码
1 #include<stdio.h> 2 int main() 3 { 4 int n; 5 while(scanf("%d",&n)!=EOF) 6 { 7 if(n%4==2) printf("yes\n"); 8 else 9 printf("no\n"); 10 } 11 return 0; 12 }
第二种方法，直接取对３取模！
由同余式的基本性质：
（1）自反性：a = a( mod m)。
以及同余式的四则运算法则：
（1）如果 a =b( mod m)且 c = d( mod m)，则 a +c = (b + d)( mod m)。
可知，F(n) = F(n) ( mod m) = ( F(n-1) +F(n-2) )( mod m)。
下面给出ＡＣ代码：
1 #include <iostream> 2 #include <stdio.h> 3 using namespace std; 4 int f[1000100]; 5 int main() 6 { 7 f[0]=1; f[1]=2; 8 int n; 9 while(scanf("%d",&n)!=EOF) 10 { 11 for(int i=2;i<=n;i++) 12 { 13 f[i] = (f[i-1]+f[i-2])%3; 14 } 15 if(f[n]%3==0) 16 printf("yes\n"); 17 else 18 printf("no\n"); 19 } 20 return 0; 21 }
