POJ3104 Drying

Drying

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 15604 Accepted: 3976

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 #132 3 95sample input #232 3 65

Sample Output

sample output #13sample output #22

Source

Northeastern Europe 2005, Northern Subregion

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题目的意思是给出n条衣服，每条衣服带水ai，单位时间每条会自然风干减少1单位水，有一台烘干机，单位时间可以帮助减少k单位水（不属于自然风干了），求全部弄干需要多少时间

思路：二分+验证 二分时间，算出能否在该时间烘干，设当前为x，每件衣服应该自然减少x单位，烘干机可视为减少（k-1）单位，然后去检查能否完成

注意：k可能等于1，可能会出现/0ＲＥ情况　特判即可

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <string>#include <vector>using namespace std;#define inf 0x3f3f3f3f#define LL long longLL a[100005],k;int n;bool ok(LL x){    LL ans=0;    for(int i=0; i<n; i++)    {        LL xx=max((LL)0,a[i]-x);        ans=ans+(xx+k-1)/k;    }    if(ans<=x)        return 1;    return 0;}int main(){    while(~scanf("%d",&n))    {        for(int i=0; i<n; i++)        {            scanf("%lld",&a[i]);        }        scanf("%lld",&k);        if(k==1)        {            LL ans =-1;            for(int i=0; i<n; i++)                ans=max(ans,a[i]);            printf("%lld\n",ans);        }        else        {            k--;            LL l=0,r=1e9;            LL ans=-1;            while(l<=r)            {                LL mid=(l+r)/2;                if(ok(mid))                {                    r=mid-1;                    ans=mid;                }                else                    l=mid+1;            }            printf("%lld\n",ans);        }    }    return 0;}