這道題,雖然是個水題,但是沒有一個數學公式,
難死了= =!,我暴力了很多遍還是沒有成功,
我知道不能用暴力破解,但實在想不出來有什麼方法,
沒辦法,求助咯,最終才知道有這麼個餘數公式:
(a+b)%3 = (a%3+b%3)%3
有了這個公式,那真是手到擒來了!
這題也展現出ACM與數學那不可分割的關系啊~
Fibonacci Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32807 Accepted Submission(s): 15874
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
#include <iostream>
using namespace std;
int fib[1000001];
int main()
{
int i,n;
fib[0]=7,fib[1]=11;
for(i=2;i<1000001;i++)
fib[i]=(fib[i-1]%3+fib[i-2]%3)%3;
while(cin>>n)
{
if(fib[n]==0)
cout<<"yes"<<endl;
else
cout<<"no"<<endl;
}
return 0;
}