HDU 5914 Triangle(规律+斐波那契数列)

Triangle

Problem Description

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≤20), which indicates the number of test cases.

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

3
4
5
6

Sample Output

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

Source

2016中国大学生程序设计竞赛（长春）-重现赛

题意：

有n根木棍，长度分别为1、2、3…n的木棍，计算至少去掉几根木棍使得剩下的木棒不能构成三角形。

思路:

三角形形成的条件：两边之和大于第三边，我们可以找出边界条件，即两边之和等于第三边，这时我们可以想到斐波那契数列。斐波那契数列的话，那么就不能组成三角形。

1 2 3 5 8 13 21

所以此题，就可以找把除斐波那契数的去掉即可。

AC代码:

#include 
using namespace std;
const int maxn = 100;
int F[maxn];
int main()
{
    int n;
    cin >> n;
    F[0] = 1;
    F[1] = 2;
    int tot = 2;
    for(int i = 2; F[i] <= 20; i++) {
        F[tot++] = F[i-1] + F[i-2];
    }
    for(int i = 1; i <= n; i++) {
        int t;
        cin >> t;
        if(t<=3) {
            printf("Case #%d: %d\n", i, 0);
        }
        else {
            int ans =  0;
            for(int j = 0; F[j] <= t; j++) {
                ans++;
            }
            printf("Case #%d: %d\n", i, t-ans);
        }
    }

    return 0;
}