HDU - 5914 A - Triangle

Triangle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1171    Accepted Submission(s): 701

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

 题目大意：有n跟木条，每根木棒的长度分别为：1.2.3.4.5......n.问最从中最少去掉几根木棒就能使得剩下的木棒不能组成三角形

思路：本题是一个斐波那契数列，另外，本题最多只有20根木条，可也直接用打表法

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int a[30]={0,0,0,0,1,1,2,3,3,4,5,6,7,7,8,9,10,11,12,13,14,15};
int main()
{
    int n;
    while(cin>>n)
    {
        for(int i=1;i<=n;i++)
        {
            int n;
            cin>>n;
            printf("Case #%d: ",i);
            cout<<a[n]<<endl;
        }
    }
}

用斐波那契数列求解（赛后参考的大神的，自己当时没想到……）

#include<stdio.h>  
#include<string.h>  
using namespace std;  
int f[50];  
void init()  
{  
    f[1]=1;  
    f[2]=2;  
    for(int i=3;i<=20;i++)  
    {  
        f[i]=f[i-1]+f[i-2];  
    }  
}  
int main()  
{  
    int t;  
    int kase=0;  
    init();  
    scanf("%d",&t);  
    while(t--)  
    {  
        int n;  
        scanf("%d",&n);  
        int tmp=0;  
        for(int i=1;i<=20;i++)  
        {  
            if(f[i]<=n)tmp++;  
        }  
        printf("Case #%d: ",++kase);  
        printf("%d\n",n-tmp);  
    }  
}