#148. 【NOIP2015】跳石头 （二分答案，最大化最小值）

题目链接
题意：略
解法：和上一道poj2456一样， 这里是删除m个点，等价于选择n-m
个点，剩下的与poj2456处理一样，经典题目

#include
#include

//#include
using namespace std;
#define LL long long
#define pb push_back
#define X first
#define Y second
#define cl(a,b) memset(a,b,sizeof(a))
typedef pair<long long ,long long > P;
const int maxn=500005;
const LL inf=1LL<<45;
const LL mod=1e9+7;

LL a[maxn];
int n,m;
LL L;
bool ok(int d){
    int last=0;
    for(int i=1;i2-m;i++){
        int cur=last+1;
        while(cur<=n+1&&a[last]+d>=a[cur])cur++;
        if(cur>n+1)return false;
        last=cur;
    }
    return true;
}
void solve(){
    int l=0,r=999999999;
    while(r-l>1){
        int mid=(l+r)/2;
        if(ok(mid))l=mid;
        else r=mid;
    }
    printf("%d\n",r);
}
int main(){
    while(~scanf("%lld%d%d",&L,&n,&m)){
        for(int i=1;i<=n;i++){
            scanf("%lld",&a[i]);
        }
        a[0]=0;a[n+1]=L;
        solve();
    }
    return 0;
}