POJ_3104_Drying_(二分,最小化最大值)

描述

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http://poj.org/problem?id=3104

n件衣服,第i件衣服里面有水a[i],自然风干每分钟干1个水,用吹风机每分钟干k个水,但是同时只能对一件衣服使用吹风机,求干完所有衣服所需时间的最小值.

Drying

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12639		Accepted: 3256

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

Source

Northeastern Europe 2005, Northern Subregion

分析

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二分.

最小化最大值.

y为原来的最大时间.然后二分可行时间x,如果有自然风干时间>x的,用吹分机加速,多出的时间用多的速度弥补,并向上取整,看吹风机是否够用.

 

注意:

1.k=1时,v=k-1=0,不能做除数.............要特判.........

#include<cstdio>
#include<algorithm>
using std :: max;

const int maxn=100005;
int n,k,x,y;
int a[maxn];

bool C(int x)
{
    int rest=x;
    for(int i=1;i<=n;i++)
    {
        if(a[i]>x)
        {
            int d=a[i]-x;
            int v=k-1;
            int res=d/v;
            if(d%v!=0) res++;
            rest-=res;
        }
        if(rest<0) return false;
    }
    return true;
}

void solve()
{
    while(x<y)
    {
        int m=x+(y-x)/2;
        if(C(m)) y=m;
        else x=m+1;
    }
    printf("%d\n",x);
}

void init()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) { scanf("%d",&a[i]); y=max(y,a[i]); }
    scanf("%d",&k);
}

int main()
{
    freopen("dry.in","r",stdin);
    freopen("dry.out","w",stdout);
    init();
    if(k==1)
    {
        printf("%d\n",y);
        return 0;
    }
    solve();
    fclose(stdin);
    fclose(stdout);
    return 0;
}