杭电1021

最新推荐文章于 2019-08-12 09:35:13 发布

fk961859482

最新推荐文章于 2019-08-12 09:35:13 发布

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分类专栏：杭电DP题库

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杭电1021

http://acm.hdu.edu.cn/showproblem.php?pid=1021

FibonacciAgain

Time Limit:2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39152 Accepted Submission(s): 18784

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, eachcontaining an integer n. (n < 1,000,000).

Output

Print the word "yes" if 3 divide evenlyinto F(n).

Print the word "no" if not.

Sample Input

Sample Output

yes

注意用如果直接模拟，会超时。这里需要借助模运算的性质：

(A+B)%C=(A%C+B%C)%C;

第一次AC的代码：

#include<iostream>
//#include<cstdio>
//#include<cstring>
//#include<cmath>
using namespace std;
int a[1000002];
int fun(int n )
{
    int i;
   a[0]=7;a[1]=11;
   for(i=2;i<=n;i++)
   a[i]=(a[i-1]%3+a[i-2]%3)%3;
   if(a[n]==0)
   return 1;
   else
   return 0;
}

int main()
{
    int n;
    while(cin>>n)
    {
        if(fun(n))
        cout<<"yes\n";
        else
        cout<<"no\n";
    }


    return 0;
}

后来发现手动输入过程中，只要是2，6,10,14,18。。。。都会输出yes，瞬间发现这题就是找规律啊，只要对4取模余2的都直接输出yes，否则输出no.

怒改代码：

#include<iostream>
void main()
{
    int n;
    while(std::cin>>n)
    std::cout<<((n%4==2)?"yes\n":"no\n");
 }