HDU5914

HDU5914
Triangle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1413 Accepted Submission(s): 840

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

Recommend
wange2014

Statistic | Submit | Discuss | Note
解析：
20以内从小往大删：
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20
1+2=3删掉2-3的数，2+3=5删掉3-5之间的数，3+5=8删掉5-8之间的数，5+8=13删掉8-13之间的数，8+13=21删掉13-21之间的数
得到：1，2，3，5，8，13，21
所以打表删除的数为：
a[30]={0,0,0,0,1,1,2,3,3,4,5,6,7,7,8,9,10,11,12,13,14};
之后直接输出就行

#include<stdio.h>int main(){    int i,n,t,a[30]={0,0,0,0,1,1,2,3,3,4,5,6,7,7,8,9,10,11,12,13,14};    scanf("%d",&t);    for(i=1;i<=t;i++)    {        scanf("%d",&n);        printf("Case #%d: %d\n",i,a[n]);     }     return 0;}

也有人用队列模拟

#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;  }