Codefroces 760 B. Frodo and pillows
1 second
256 megabytes
standard input
standard output
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?
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.
Print single integer — the maximum number of pillows Frodo can have so that no one is hurt.
4 6 2
2
3 10 3
4
3 6 1
3
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.
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int main()
{
ll count,n,m,k,ans,pos,cnt,inf;
while(scanf("%I64d%I64d%I64d",&n,&m,&k)!=EOF)
{
cnt=;count=;inf=;
m-=n;ans=;
pos=max(k-,n-k);
for(ll i=;i<=pos;i++)
{
cnt+=(i*-);//交替铺最多消耗枕头数
}
if(m>cnt)//平铺
{
inf=(m-cnt)/n;
m-=(inf*n);
}
while(m>)//交替铺
{
if(k-count>=) ans++;
if(k+count<=n) ans++;
m-=ans;
count++;
}
printf("%I64d\n",inf+count);
}
return ;
}
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