题目
n hobbits are planning to spend the night at Frodo's house. Frodo has n beds standing in a row and m pillows (n ≤ m). Each hobbit needs a bed and at least one pillow to sleep, however, everyone wants as many pillows as possible. Of course, it's not always possible to share pillows equally, but any hobbit gets hurt if he has at least two pillows less than some of his neighbors have.
Frodo will sleep on the k-th bed in the row. What is the maximum number of pillows he can have so that every hobbit has at least one pillow, every pillow is given to some hobbit and no one is hurt?
Input
The only line contain three integers n, m and k (1 ≤ n ≤ m ≤ 109, 1 ≤ k ≤ n) — the number of hobbits, the number of pillows and the number of Frodo's bed.
Output
Print single integer — the maximum number of pillows Frodo can have so that no one is hurt.
Examples
Input
4 6 2
Output
2
Input
3 10 3
Output
4
Input
3 6 1
Output
3
Note
In the first example Frodo can have at most two pillows. In this case, he can give two pillows to the hobbit on the first bed, and one pillow to each of the hobbits on the third and the fourth beds.
In the second example Frodo can take at most four pillows, giving three pillows to each of the others.
In the third example Frodo can take three pillows, giving two pillows to the hobbit in the middle and one pillow to the hobbit on the third bed.
题目大意
n个霍比特人排成一行准备睡觉，他们有n张床和m个枕头，弗拉多作为主人要给这些人分枕头。每个霍比特人都希望自己有尽可能多的枕头，而每个霍比特人只要比跟他相邻的人少一个以上的枕头（不包括一个），他就会受到伤害，问弗拉多在不会伤害任意一个人的情况下能获得最多多少枕头。
算法：二分 贪心
代码
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
long long n,m,k;
cin>>n>>m>>k;
long long low=1,high=m,mid=(low+high)/2;
while(low<=high)
{
long long sum=n+mid-1;
if(mid-(k-1)>=1) sum+=(2*mid-k-2)*(k-1)/2;
else sum+=(mid-2)*(mid-1)/2;
if(mid-(n-k)>=1) sum+=(2*mid-3-n+k)*(n-k)/2;
else sum+=(mid-2)*(mid-1)/2;
if(sum>m) {high=mid-1;mid=(high+low)/2;}
else {low=mid+1;mid=(high+low)/2;}
}
cout<<mid;
return 0;
}
代码解析
从k这个位置，左右阶梯状散发，但是左右有可能很快就到头了，所以需要判断
代码中的第一个if else就是用来判断左端并且根据情况计算
相应的第二个if else是判断右边