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HDOJ1021:Fibonacci Again——简单数学题

HDOJ1021:Fibonacci Again——简单数学题

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 判断能否被3整除。可以打表做,简单AC。 代码:

#include
using namespace std;
int main(){
   int n;
   while(cin>>n){
        if(n%4==2) cout<<"yes"<
                
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HDOJ1021:Fibonacci Again——简单数学题
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 conta…
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2015-03-18