POJ 3104 Drying

Posted 2022-02-23 cq.tiancx

题目： POJ 3104 Drying ，哈哈，我们今天来看一道稍微复杂一点的二分答案的题嘛，这是选自POJ上的一道题，好了，我们一起来看看题意吧：

题目描述是复制的，可能有部分显示不对，我就把题目链接放下面！
题目链接： POJ 3104 Drying

题目描述

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.

输入描述

The first line contains a single integer n (1 ≤ n ≤ 100 000). The second line contains ai separated by spaces (1 ≤ ai ≤ 10^9). The third line contains k (1 ≤ k ≤ 10^9).

输出描述

Output a single integer — the minimal possible number of minutes required to dry all clothes.

示例1

输入

3
2 3 9
5

输出

3

示例2

输入

3
2 3 6
5

输出

2

思路:

这道题我们采用二分答案的思路去做！具体的直接看代码，有注释！

我们来看看成功AC的代码吧：

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;//好像不开long long 也能AC
int n,k;
int a[100010];
ll l,r;
int check(int x)
    int t=0;
    for(int i=1;i<=n;i++)
        if(a[i]<=x) continue;
        t+=(int)ceil((double)(a[i]-x)*1.0/(k-1));//a[i]-x)*1.0/(k-1) 表示该件衣服的最少用时
        if(t>x) return 0;//若用时大于x，则说明时间短了
    
    return 1;

int main()
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    while(cin>>n)
        int mmax=0;
        for(int i=1;i<=n;i++) cin>>a[i],mmax=max(mmax,a[i]);
        sort(a+1,a+1+n);
        cin>>k;
        if(k==1) cout<<mmax<<"\\n"; continue;//这里需要特判下
        l=0,r=mmax;
        while(l<r)//二分模板
            ll mid=(l+r)>>1;
            if(check(mid)) r=mid;
            else l=mid+1;
        
        cout<<l<<"\\n";
    

    return 0;

谢谢你的阅读，由于作者水平有限，难免有不足之处，若读者发现问题，还请批评，在留言区留言或者私信告知，我一定会尽快修改的。若各位大佬有什么好的解法，或者有意义的解法都可以在评论区展示额，万分谢谢。
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