poj3258

#include 
#include 
#include 

using namespace std;

long long rock[50005],l,n,m;

bool check(long long d)
{
    int total=0,i=1,j=1;
    while (i<n)
    {
        j=i+1;
        while (rock[j]-rock[i]<d&&j<n)
        {
            total++;
            j++;
        }
        i=j;
    }
    if (rock[n]-rock[n-1]<d) total++;
    return total<=m;
}

int main()
{
   // freopen("in.txt","r",stdin);
    cin>>l>>n>>m;
    n+=2;
    for (int i=2;i<=n-1;i++) cin>>rock[i];
    rock[n]=l;
    sort(rock+1,rock+n+1);
    long long left=1,right=rock[n],mid;
    while (right-left>1)
    {
        mid=(left+right)/2;
        //cout<
        if (check(mid)) left=mid;else right=mid;
    }
    if (check(right)) cout<<right;else cout<<left;
    return 0;
}

题意：一些石头排成一条线，第一个和最后一个不能去掉，其余的共可以去掉m块，要使去掉后石头间距的最小值最大。