HDU 1021 Fibonacci Again

　　　　　　　　　　　　HDU 1021 Fibonacci Again

　　　　　　Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
　　　　　　　　　　　Total Submission(s): 68704    Accepted Submission(s): 31604

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

 

思路：f(0) = 7, f(1) = 11, f(n) = f(n-1) + f(n-2)    童鞋们应该很快就想到了题目中的 Fibonacci

然而水平非常low的我只会写递归，so 。。MLE       then我发现 n % 4 == 2 输出“yes”，否则输出“no”

然后我去找了参加省选的学姐（dalao一枚），she said :"你用dfs当然会MLE了，你改成用递推，你递推会写吗？"

然后我告诉了她我找的规律，然后学姐默默地找到了另一个规律：1 2 0 2 2 1 0 1 1 2 0 2 2 1 0 1 ......

但是我还是按我自己的写了（比较lan）

所以一共有三种写法

#include<algorithm>
#include<cstdio>
using namespace std;
int n;
int f[1000005];

int dfs(int x) {
    if(f[x] != 0) return f[x];
    f[x] = dfs(x-1) + dfs(x-2);
}

int main() {
    f[0] = 7; f[1] = 11;
    while(scanf("%d", &n) != EOF) {
        int t = dfs(n);
        if(t%3 == 0) printf("yes
");
        else printf("no
");
    }
    return 0;
}

#include<cstdio>
using namespace std;
int n;

void judge(int x) {
    if(x%4 == 2) printf("yes
");
    else printf("no
"); 
}

int main() {
    while(scanf("%d", &n) != EOF) {
        judge(n);
    }
    return 0;
}

其他你们就自己写吧  嗯