POJ 3104 Drying（二分

Drying

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 22163		Accepted: 5611

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 

題意:有n件衣服，每件衣服有一個溼度ai，若是把衣服放着，這些衣服每秒鐘會天然風乾一點適度，若是把衣服放在熱機上面每秒鐘會熱幹k-1點溼度值嗎，求最少多少秒會風乾

題解：二分時間，每次枚舉的時間後，將原先的溼度值減小改時間的溼度值，而後將每一個大於（k-1）溼度值的放在熱機上計算所須要的時間，由於時間是線性單調的，因此就能夠使用二分來求

代碼以下：

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]";
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int maxn = 1e5+5;
/*
> 二分枚舉時間,判斷可行性,而後求解便可.
> 
> 注意判斷可行性的過程:
> 
> 1.先把全部衣服的含水量減去T
> 2.而後把>=(k-1)的拿去烘乾，
能夠理解爲烘乾時候每分鐘掉(k-1)水，
這樣全部的衣服都每分鐘天然幹掉1水了。
由於每分鐘掉一滴水是確定的了，
所以，若是你去烘乾它的話，
那麼它就能再掉多k-1 + 1 == k，
這樣纔是k滴水,而後就是計算總花費時間
> 3.最後判斷總花費時間時候小於T,
若小於等於,則可行性,不然不可行.
*/
LL a[maxn];
LL b[maxn];
int main(){

    LL n,k;
    int cas=1;
    while(~scanf("%lld",&n)){
        for(int i=0;i<n;i++){
            scanf("%lld",&a[i]);
        }
        LL l=0;
        LL r=1e9;
        scanf("%lld",&k);
        if(k==1){
            LL ans=0;
            for(int i=0;i<n;i++){
                ans=max(ans,a[i]);
            }
            cout<<ans<<endl;

        }else{
            while(l<r){
                LL mid=(l+r)/2;
                for(int i=0;i<n;i++) b[i]=a[i]-mid;
                    LL t=0;
                for(int i=0;i<n;i++){
                    if(b[i]>=(k-1)){
                        if(b[i]%(k-1)==0) t+=b[i]/(k-1);
                        else{
                            t+=b[i]/(k-1) + 1;
                        }
                    }else if(b[i]>0){
                        t++;
                    }
                }
                if(t<=mid) r=mid;
                else l=mid+1;
            }
            cout<<r<<endl;
        }
    }
}   

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