HDU 5914 Triangle【斐波那契数列】

Triangle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 178    Accepted Submission(s): 127

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

3456

 

Sample Output

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

 

Source

2016中国大学生程序设计竞赛（长春）-重现赛

题意：有1~n n个数，问最少拿走多少个数能使剩下的数不能构成三角形。

规律:

     首先考虑一下，除开123不谈，任意三个相邻的数是肯定可以组成三角形的；并且如果有两个数很近的时候，也肯定可以和另一个数组成三角形（两边之差小于第三边）；因此我们考虑有一定间隔的情况，即对于剩下的边总有a[i]+a[i+1]=a[i+2]，因此想到斐波那契数列，接下来我们证明它的充要性；

充分性：a[i]+a[i+1]=a[i+2]，那个对于每一个比a[i+2]更大的数一定小于a[i]+a[i+1]，因此可证；

必要性：如果有任意一个比a[i+1]大一点的数，那么要么对于当前的a[i+2]满足两边之和大于第三边，就可以构成三角形；如果是小一点，那么对于前面的一组数又可以组成三角形，因此可证；

因此斐波那契数列可行~

那么就需符合：1 2 3 5 8 13......

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<climits>#include<string>#include<queue>#include<stack>#include<set>#include<map>#include<algorithm>using namespace std;#define rep(i,j,k)for(i=j;i<k;i++)#define per(i,j,k)for(i=j;i>k;i--)#define MS(x,y)memset(x,y,sizeof(x))#define max(a,b) a>b?a:b#define min(a,b) a<b?a:b#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef long long LL;const int INF=0x7ffffff;int num[22]={0,0,0,0,1,1,2,3,3,4,5,6,7,7,8,9,10,11,12,13,14};int main(){  int T;  int Case=1;  int n;  scanf("%d",&T);  while(T--){    scanf("%d",&n);    printf("Case #%d: %d\n",Case++,num[n]);  }  return 0;}