HDU-5914 Triangle（思路）

最新推荐文章于 2024-08-29 12:01:12 发布

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最新推荐文章于 2024-08-29 12:01:12 发布

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分类专栏：模拟构造规律二分三分文章标签：迭代思维

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探讨了在1到n的整数集合中，为了防止任意三数构成三角形，需要移除的最少整数数量。通过算法实现，采用两变量作为边长，遍历第三边，若大于两变量之和则移除。

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Triangle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 203 Accepted Submission(s): 145

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

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：1，2……n共n个数中，至少抽出多少数才可以不可能组成三角形。

思路：两个变量充当两边，枚举第三个数，如果大于两个变量的和就去掉第三个数，依次迭代。

ps：还看到有人用的什么去掉不是斐波那契数的就可以了，这个是为什么？是有什么定理，还是靠规律和结果推得啊，求大牛随手解答。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int main()
{
    int t,n;
    cin >> t;
    for(int ca = 1;ca <= t;ca++)
    {
        scanf("%d",&n);
        printf("Case #%d: ",ca);
        if(n <= 3)
            printf("0\n");
        else
        {
            int a = 2,b = 3,sum = 0;
            for(int i = 4;i <= n;i++)
            {
                if(a+b > i)
                    sum++;
                else
                    a = b,b = i;
            }
            printf("%d\n",sum);
        }
    }
    return 0;
}