本文主要是介绍HDU_1021 Fibonacci again (水题),希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
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
解题思路:
第一眼看题觉得是个大数处理问题 , 后来写了几个例子发现是单纯的规律题。
废话不多说,直接放代码。
代码:
#include <stdio.h>
using namespace std;int main(){long n;while(scanf("%ld",&n) != EOF){if(n == 2 || (n-2)%4 == 0)printf("yes\n");elseprintf("no\n");}return 0;
}
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