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HDU1021

时间:2017-12-07 21:03:12      阅读:123      评论:0      收藏:0      [点我收藏+]

标签:word   sam   int   using   into   include   des   input   text   

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 65456    Accepted Submission(s): 30363


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
 
= =。。。一开始用递归,内存超限。。。随后用这个递归得出规律,就是从第二个开始每隔4个就一次YES。。。AC
 
 1 #include <iostream>
 2 using namespace std;
 3 long FIB(long n);
 4 int main()
 5 {
 6     long n;
 7     while(cin>>n)
 8     {
 9         if(FIB(n)%3==0)
10             cout<<"yes"<<endl;
11         else
12             cout<<"no"<<endl;
13     }
14 
15     return 0;
16 }
17 
18 long FIB(long n)
19 {
20     if(n==0)  return 7;
21     if(n==1)  return 11;
22     return FIB(n-1)+FIB(n-2);
23 }

找到规律后:

 1 #include <iostream>
 2 using namespace std;
 3 long FIB(long n);
 4 int main()
 5 {
 6     long n;
 7     while(cin>>n)
 8     {
 9         if(n%4==2)
10             cout<<"yes\n";
11         else
12             cout<<"no\n";
13     }
14 
15     return 0;
16 }

 

HDU1021

标签:word   sam   int   using   into   include   des   input   text   

原文地址:http://www.cnblogs.com/BOW1203/p/8000641.html

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