Triangle

最新推荐文章于 2018-11-04 15:46:40 发布

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最新推荐文章于 2018-11-04 15:46:40 发布

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分类专栏： ACM STL

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本篇博客介绍了一个有趣的算法问题：如何确定最少需要移除多少根长度递增的棍子，以确保剩下的棍子无法组成任何三角形。通过队列模拟的方法实现了这一目标，并提供了完整的C++代码实现。

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

我是用队列模拟写的。


#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int main()
{
    queue<int>que;
    int a[100];
    a[0]=a[1]=a[2]=a[3]=0;
    que.push(3);
    que.push(2);
    int n,m,k;
    int ans=0;
    k=4;
    while(k<=20)
    {
        n=que.front();
        que.pop();
        m=que.front();
        if(n+m>k)
        {
            ans++;
            que.push(n);
            a[k]=ans;
        }
        else
        {
            que.push(k);
            a[k]=ans;
        }
        k++;
    }
    int p=1;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&m);
        printf("Case #%d: ",p);
        printf("%d\n",a[m]);
        p++;
    }
    return 0;
}