这个二分查找好难（Drying）

Drying

It is very hard to wash and especially to dry clothes in winter. But Jane is a very smart girl. She is not afraid of this boring process. Jane has decided to use a radiator to make drying faster. But the radiator is small, so it can hold only one thing at a time.

Jane wants to perform drying in the minimal possible time. She asked you to write a program that will calculate the minimal time for a given set of clothes.

There are n clothes Jane has just washed. Each of them took ai water during washing. Every minute the amount of water contained in each thing decreases by one (of course, only if the thing is not completely dry yet). When amount of water contained becomes zero the cloth becomes dry and is ready to be packed.

Every minute Jane can select one thing to dry on the radiator. The radiator is very hot, so the amount of water in this thing decreases by k this minute (but not less than zero — if the thing contains less than k water, the resulting amount of water will be zero).

The task is to minimize the total time of drying by means of using the radiator effectively. The drying process ends when all the clothes are dry.

Input

The first line contains a single integer n (1 ≤ n ≤ 100 000). The second line contains ai separated by spaces (1 ≤ ai ≤ 109). The third line contains k (1 ≤ k ≤ 109).

Output

Output a single integer — the minimal possible number of minutes required to dry all clothes.

Sample Input

3
2 3 9
5
3
2 3 6
5

Sample Output

3
2

大概意思是：第一行为N，第二行N个数Ai，第三行为K。每次挑选一个Ai减去K-1，然后每个Ai减去1，问几次操作才能使所有Ai小于等于0.打死都不会呀！。。。做题链接 https://vjudge.net/contest/237547#problem/D

思路就是二分次数。找到一个次数t，使得小于t的数都能只通过减1符合题意，大于这个次数的数恰好总共减t次d和各减一次t从而符合题意。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
using namespace std;
long long n,d;
const int MAX_N = 100005;
int clo[MAX_N];
int check(long long p){
    long long num = 0;
    for(int i = 1;i<=n;++i)
        if(p>=clo[i]) continue;
        else 
            num+=(((clo[i]-p-1)/(d-1))+1);
    if(num<=p) return 1;
    else return 0;
}
int main(){
    int maxx;
    while(scanf("%lld",&n)==1){
            maxx = -1;
     for(int i= 1;i<=n;++i){
            scanf("%d",&clo[i]);
              if(maxx<clo[i]) maxx=clo[i];
     }
     scanf("%lld",&d);
     if(d==1) printf("%d\n",maxx);
     else {
        int l=0,r=maxx;
        while(l<=r){
        //dbg(l);
        //dbg(r);
            int mid=(l+r)>>1;
            if(check(mid)) r = mid-1;
            else l = mid+1;
        }
       printf("%d\n",l);
     }
    }
    return 0;
}