HDU 1021 斐波那契

参考自：https://www.cnblogs.com/ECJTUACM-873284962/p/6404504.html

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.

 

Sample Input

0

1

2

3

4

5

Sample Output

no

no

yes

no

no

no

 

解法：

#include<stdio.h>
int main(){
    int i,n,f[10000];
    f[0] = 7;
    f[1] = 11;
    while(scanf("%d",&n)!=EOF){
        for(i=2; i<=n; i++){
            f[i] = f[i-1]+f[i-2];
        }
        if(f[n]%3 == 0)
            printf("yes
");
        else printf("no
");
    }
    return 0;
}

后来，get一个规律：n除4余2的就输出yes，否则no。

#include<stdio.h>
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        if(n%4==2)
            printf("yes
");
        else printf("no
");
    }
    return 0;
}