codeforces E. Convention(二分)

E. Convention

time limit per test:1 second
memory limit per test:256 megabytes
input:standard input
output:standard output

Cows from all over the world are arriving at the local airport to attend the convention and eat grass. Specifically, there are N cows arriving at the airport (1≤N≤10⁵) and cow ii arrives at time titi (0≤ti≤10⁹). Farmer John has arranged M (1≤M≤10⁵) buses to transport the cows from the airport. Each bus can hold up to C cows in it (1≤C≤N). Farmer John is waiting with the buses at the airport and would like to assign the arriving cows to the buses. A bus can leave at the time when the last cow on it arrives. Farmer John wants to be a good host and so does not want to keep the arriving cows waiting at the airport too long. What is the smallest possible value of the maximum waiting time of any one arriving cow if Farmer John coordinates his buses optimally? A cow’s waiting time is the difference between her arrival time and the departure of her assigned bus.

It is guaranteed that MC≥N.

Input

The first line contains three space separated integers N, M, and C. The next line contains N space separated integers representing the arrival time of each cow.

Output

Please write one line containing the optimal minimum maximum waiting time for any one arriving cow.

Example

input

6 3 2
1 1 10 14 4 3

output

4

分析

牛的等待时间的数据很大，而且单一，可以想到用二分法；
直接二分等待时间(0~10⁹)，排序后塞牛进车；如果车辆不够用则证明太小，缩小左区间，否则缩小右区间，直到二分完成。

代码

#include
#include
#include
#include
#include
#define MS(X) memset(X,0,sizeof(X))
#define MSC(X) memset(X,-1,sizeof(X))
typedef long long LL;
using namespace std;
const int maxt=1e9;
int a[100005],n,m,c;
int main(){
    int lb=0,ub=maxt,pt,mchk,cchk,fnt,fi;
    scanf("%d%d%d",&n,&m,&c);
    for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
    sort(a,a+n);
    while(ub-lb>1){
        mchk=m;cchk=1;
        int i;
        pt=(lb+ub)/2;
        fnt=a[0];
        for(i=1;i<n&&mchk;i++){
            if(a[i]-fnt>pt||cchk>=c){
                mchk--;
                fnt=a[i];
                cchk=1;
            }
            else if(cchk<c)
                cchk++;
        }
        if(mchk==0) lb=pt;
        else ub=pt;
    }
    printf("%d\n",ub);
    return 0;
}