POJ 3104

Drying

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 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.

题目大意

给你一些带有水衣服 需要干燥
干燥方法有自然风干每单位时间风干一份水 用烘干机每次一件烘干 k 份水
让你求出干燥全部衣服的最少用时

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

分析

时间最多为 水分最多衣服的风干时间
使用二分来查找最少时间 用 mid 代表烘干机用的时间 小于 mid 则忽略
每件使用烘干机的时间为 t 则 k*t=p[i]-(mid-t)
则每件时间为t=(p[i]-mid)/(k-1)

#include<stdio.h>
#include<algorithm>
using namespace std;
long long i,n,k,a,b,c,mid,maxx,p[101010];
int main()
{
	scanf("%lld",&n);
	for(i=0;i<n;i++)
		scanf("%lld",&p[i]);
	scanf("%lld",&k);
	sort(p,p+n);
	if(k==1) printf("%lld\n",p[n-1]);
	else
	{
		a=0,b=p[n-1];
		while(b>a+1)
		{
			c=0;
			mid=(a+b)/2;
			for(i=0;i<n;i++)					//k*t=p[i]-(mid-t)
				if(p[i]>mid)					//t=(p[i]-mid)/(k-1)
					c+=(p[i]-mid+(k-1)-1)/(k-1);//取整方法:分子后面+((分母)-1)
			if(c>mid) a=mid;
			else b=mid;
		}
		printf("%lld\n",b);
	}
}