HDU 1020-1029

1020

1021 Fibonacci again

  问题重述:F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2);给定一系列n,求f(n)是否被3整除;

  思路:硬算,暴力求解;

  正确思路:找规律,发现n%4==2那么就可以被f(n)就可以被3整除;

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 using namespace std;
 6 int main() {
 7     int n;
 8     while(~scanf("%d",&n)){
 9         if(n%4==2)puts("yes");
10         else puts("no");
11     }
12     return 0;
13 }
1021 规律解

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转载自www.cnblogs.com/yidiandianwy/p/11593578.html