POJ 3104 Drying(二分答案)

Drying

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24089		Accepted: 6031

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 kwater, 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

Source

Northeastern Europe 2005, Northern Subregion

【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

【分析】

二分答案。但需注意：

①每分钟烘干k单位的水，于是我就想当然地除k向上取整了((a_i– mid) / k)。其实应该除以k-1，列个详细的算式：

设需要用x分钟的机器，那么自然风干需要mid – x分钟，x和mid需要满足：

k*x + (mid – x) >=a_i,即 x >= (a_i– mid) / (k – 1)。

②当k=1的时候，很显然会发生除零错误，需要特殊处理

【代码】

#include<cstdio>
#include<algorithm>
#include<iostream>
#define debug(x) cerr<<#x<<" "<<x<<'\n';
using namespace std;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int N=1e5+5;
int ans,n,k,d[N];
inline bool check(int now){
	int res=0;
	for(int i=1;i<=n;i++){
		int left=d[i]-now;
		if(left>0){
			res+=(left+k-1)/k;
			if(res>now) return 0;
		}
	}	
	return 1;
}
int main(){
	n=read();
	for(int i=1;i<=n;i++) d[i]=read();
	k=read();k--;
	sort(d+1,d+n+1);
	if(!k){printf("%d\n",d[n]);return 0;}
	int l=0,r=d[n],mid;
	while(l<=r){
		mid=l+r>>1;
		if(check(mid)){
			ans=mid;
			r=mid-1;
		}
		else{
			l=mid+1;
		}
	}
	printf("%d\n",ans);
	return 0;
}

POJ 3104 Drying(二分答案)

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