文章标题 SPOJ PHT ： Pigeonhole Tower（二分）

Pigeonhole Tower

Pigeon SSNA want to build a tower with some wood walls. Let’s describe the tower they want to make:

A Tower can consist of different number of level.

If a tower contain L levels then 1st level must contain L holes , 2nd level L-1 , 3rd level L-2 ….. L level contain 1 hole .

Each room contain 3 wood walls.

See the picture below:

3 level 4level

3 Level Tower 4 Level tower

Now pigeon SSNA has n wood walls. What is maximum number of level he can made.

Input

Input starts with an integer T (≤ 100000), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 1012)

Output

For each case of input you have to print the case number and expected answer.

Sample Input

Output for Sample Input

2

15

24

Case 1: 3

Case 2: 4

题意： 给你n个木棒，然后看能堆成几层。

分析：通过看规律可知每层的数目是一个等差数列，所以对于所给数目n可以通过二分查找n所在的区间，通过所给数据可知，层数在1~1*10^6之间。

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue>
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
long long n;
int bs(){
long long lo=1,hi=10e6;
long long mid;
while (lo<hi){
mid=(hi+lo)/2;
//如果找到区间就直接返回
if (n<mid*(mid+2)&&n>=(mid-1)*(mid+1)) return mid-1;
if (n>=mid*(mid+2))lo=mid;
else hi=mid;
}
}
int main ()
{
int t;
scanf ("%d",&t);
int cnt=1;
while (t--){
scanf ("%lld",&n);
if (n<3){
printf ("Case %d: 0\n",cnt++);
continue;
}
int ans=bs();
printf ("Case %d: %d\n",cnt++,ans);
}
return 0;
}