ZJU 2060 解题报告

题目如下：

Fibonacci Again

---

Time limit: 1 Seconds   Memory limit: 32768K  
Total Submit: 3840   Accepted Submit: 1544  

---

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

Author: Leojay

---

Problem Source: ZOJ Monthly, December 2003

 

（题目地址：http://acm.zju.edu.cn/show_problem.php?pid=2060）

 

刚看题目，觉得很简单，一个递归就可以了：

 

#include <stdio.h>
#include <stdbool.h>

int F(int);
bool J(int);

int main(int argc, char *argv[])
{
    int n;
    while (scanf("%d", &n)) (J(F(n))) ? puts("yes") : puts("no");
          
    return 0;
}

int F(int n)
{
    if (n == 0) return 7;
    if (n == 1) return 11;
    return F(n - 1) + F(n - 2);
}

bool J(int f)
{
    return ((f % 3) ? false : true);
}

提交，结果提示：Time Limit Exceeded

看来程序效率较低，把递归改为非递归，如下：

#include <stdio.h>
#include <stdbool.h>

int F(int);
bool J(int);

int main(int argc, char *argv[])
{
    int n;
    while (scanf("%d", &n)) (J(F(n))) ? puts("yes") : puts("no");
          
    return 0;
}

int F(int n)
{
    int i, temp[2] = {7, 11};
    if (n == 0) return 7;
    if (n == 1) return 11;
    for (i = 2; i <= n; i ++)
        if (i % 2) temp[1] = temp[0] + temp[1];
        else temp[0] = temp[0] + temp[1];
    return temp[n % 2];
}

bool J(int f)
{
    return ((f % 3) ? false : true);
}

再次提交，还是没有AC，于是结果前20个打印出来，看是否有规律，如下：

no
no
yes
no
no
no
yes
no
no
no
yes
no
no
no
yes
no
no
no
yes
no

可见每隔3个"no"有一个"yes"，那么是不是每当 n == 4k + 2 (k >= 0) 时都有 F(n) == 3x (x >= 0) 呢？用数学归纳法试证明之：

显然 k == 0 时假设成立。设 k == m - 1 (m > 0) 时成立，则有 F(k) == F(m - 1) == F(4m - 2) == 3x(x > 0)；设 F(4m - 1) == t，则F(4m) == 3x + t；F(4m + 1) == 3x + 2t；F(4m + 2) == 6x + 3t == 3y；可知 k == m 时假设也成立。

于是写代码如下：

#include <stdio.h>

int main(int argc, char *argv[])
{
    int n;

    while (scanf("%d", &n) != EOF) (n % 4 == 2) ? puts("yes") : puts("no");
    
    return 0;
}

提交，Accepted