Triangle HDU - 5914 （斐波那契数列的应用1123）

最新推荐文章于 2018-11-04 15:46:40 发布

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本文探讨了一个有趣的数学问题——如何通过最少次数的偷取，确保无法用剩余的棒子组成三角形。给出了详细的输入输出样例及代码实现。

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Mr. Frog has n sticks, whose lengths are 1,2, 3⋯⋯n respectively. Wallice is a bad man, so he does not want Mr. Frog to form a triangle with three of the sticks here. He decides to steal some sticks! Output the minimal number of sticks he should steal so that Mr. Frog cannot form a triangle with
any three of the remaining sticks.

Input

The first line contains only one integer T (T≤20T≤20), which indicates the number of test cases.

For each test case, there is only one line describing the given integer n (1≤n≤201≤n≤20).

Output

For each test case, output one line “Case #x: y”, where x is the case number (starting from 1), y is the minimal number of sticks Wallice should steal.

Sample Input

Sample Output

Case #1: 1
Case #2: 1
Case #3: 2

刚开始看有点蒙，后来猜了几组数据，看着眼熟，于是搞出来了；

上代码;

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int  T;
int n;
int vis[20];
int main()
{
   scanf("%d", &T);
   vis[1] = vis[0] = 1;
   for(int i=2; i<=20; i++)
   {
	  vis[i] = vis[i-1] + vis[i-2];//构建斐波那契
   }
   for(int w=1; w<=T; w++)
   {
	 scanf("%d", &n);
     printf("Case #%d: ",w);
     for(int i=1; i<=20; i++)
     {
			if(vis[i] > n){cout << n-i+1 << endl;break;}
	 }
     
	}
     
	return 0;
}

水波.