POJ 3104 Drying-优快云博客

本文链接：https://blog.youkuaiyun.com/albertluf/article/details/79307367

本文介绍了一个使用烘干机来加速衣物干燥过程的算法。通过优化烘干机的使用方式，该算法能够在最短的时间内将所有衣物干燥完毕。文章还提供了一段实现此算法的C++代码。

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Drying

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions:19847		Accepted: 5008

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

不用烘干机速度为1，用了为k,提高了k-1,设要在t时间内完成，则
a[i]-t>0则说明需要用烘干机(a[i]-t)/(k-1)时。
用二分找到最小的t

代码：

#include<cstdio>
#include<algorithm>
using namespace std;
int a[100004];
int main()
{
    int n,i,j,k,ma=0;
    scanf("%d",&n);
    for(i=0; i<n; i++)
    {
        scanf("%d",&a[i]);
        ma=max(ma,a[i]);
    }
    scanf("%d",&k);
    k--;
    if(k==0)
    {
        printf("%d\n",ma);
        return 0;
    }
    int q=0,r=1000000000;
    while(r-q>1)
    {
        int mid=(r+q)/2;
        int time=0,s=0;
        for(i=0; i<n; i++)
        {
            int t=a[i]-mid;
            if(t<=0)continue;
            if(t%k)time=t/k+1;
            else time=t/k;
            s+=time;
            if(s>mid)break;
        }
        if(s>mid)q=mid;
        else if(s<=mid)r=mid;
        else q=mid;
    }
    printf("%d\n",r);
    return 0;
}