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
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
有n件衣服，其含水量给出，不做处理单位时间可减少一含水量，有且仅有一个烘干机可以单位时间减少k含水量，求将所有衣服弄干的最短时间，用二分法。对于每一个中间值，枚举每个衣服，小于x则自然风干，大于x能用烘衣机一段时间再自然风干使其时间等于x。
#include<iostream>#include<string.h>#include<cmath>
using namespace std;
long long a[100005];
int main()
{ int n; while(cin>>n) { long long max=-1; memset(a,0,sizeof(a)); for(int i=0;i<n;i++) { cin>>a[i]; if(a[i]>max) max=a[i]; } long long k; cin>>k; long long low=1,hign=max,mid; if(k!=1) {
while(hign-low>1) { mid=(low+hign)/2; long long time=0; for(int i=0;i<n;i++) { if(a[i]>mid) time=time+(a[i]-mid +k-2)/ (k-1); } if(time>mid) low=mid+1; else hign=mid; }
mid=(low+hign)/2; cout<<mid<<endl; } else cout<<max<<endl; }
return 0;}