Drying(贪心)

Drying

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11512		Accepted: 2977

Description

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 a_i 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 a_i separated by spaces (1 ≤ a_i ≤ 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

sample input #1
3
2 3 9
5

sample input #2
3
2 3 6
5

Sample Output

sample output #1
3

sample output #2
2
题解：
题意：有一些衣服，每件衣服有一定水量，有一个烘干机，每次可以烘一件衣服，每分钟可以烘掉k滴水。
每件衣服没分钟可以自动蒸发掉一滴水，用烘干机烘衣服时不蒸发。问最少需要多少时间能烘干所有的衣服。
类似于疯牛那道题；刚开始自己想的太简单了，本来自己想的//if(x+x*K>=sum)return true;直接根据水量来判断，果断wa；最后看了大神的，是根据
洗衣机时间与总时间比较的；x1+x2=x,x1+k*x2=a[i],x-x2+k*x2=a[i];
求出洗衣机时间x2；由于xiyiji没有LL，wa了n次；
ac代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
const double IEX=0.9999999999999999;
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
#define T_T while(T--)
#define mem(x,y) memset(x,y,sizeof(x))
#define SL(x) scanf("%lld",&x)
const int MAXN=100010;
typedef long long LL;
LL N,K;
LL sum;
LL a[MAXN];
bool js(LL x){
    //if(x+x*K>=sum)return true;
    //else return false;
    LL xiyiji=0;//LL;
    if(K==1)return false;
    for(int i=0;i<N;i++){
        if(a[i]<=x)continue;
        //这个x是总时间，也就是x1+x2=x,x1+k*x2=a[i],x-x2+k*x2=a[i];
        //所以要除以k-1; 
        xiyiji+=(LL)((a[i]-x+K-2)/(K-1));
        if(xiyiji>x)return false;
    }
    return true;
}
LL erfen(LL l,LL r){
    LL mid;
    while(l<=r){//
        mid=(l+r)/2;
        if(js(mid)){
            r=mid-1;
        }
        else l=mid+1;
    }
    return l;
}
int main(){
    while(~scanf("%lld",&N)){
        sum=0;
        for(int i=0;i<N;i++)SL(a[i]),sum=max(sum,a[i]);
        SI(K);
        if(K==1)printf("%lld
",sum);
        else printf("%lld
",erfen(0,sum));
    }
    return 0;
}