（POJ

（POJ - 3104）Drying

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 17840 Accepted: 4484

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 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.

Input

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

题目大意：有n件衣服，每件衣服的含水量为ai单位，每分钟他们能自然脱水1单位，有一个脱水机，每次只能对一件衣服脱水，脱水量为每分钟k单位。（脱水时不自然风干）问所有衣服全部含水量为0的最小时间。

思路：我们容易想到二分脱水的时间（这个时间介于1和最大含水量之间），对于某次二分出来的值mid：
1.对于含水量ai<=mid的衣服，直接自然风干即可；
2.对于含水量ai>mid的衣服，最少的用时是用脱水机一段时间，自然风干一段时间，设这两段时间分别为x1，x2。那么有①x1+x2=mid，②ai<=k∗x1+x2。联立二式可得：x1>=(ai−mid)/(k−1)，对这个式子向上取整就是脱干的最少用时,（用于二分的判断）。其他细节请看代码。

#include<cstdio>#include<cmath>using namespace std;typedef long long LL; const int maxn=100005;int a[maxn];int n,k;bool check(int x){    LL cnt=0;//这里必须用long long，不然会爆int,我就是在这里一直wa     for(int i=0;i<n;i++)    {         if(a[i]>x) cnt+=ceil(double(a[i]-x)/(k-1));//整数的除法会舍去小数，所以要强转double     }    if(cnt<=x) return true;//说明答案偏大了     else return false;}int main(){    while(scanf("%d",&n)!=EOF)    {        LL lo=1,hi=0,mid;        for(int i=0;i<n;i++)        {            scanf("%d",a+i);            if(hi<a[i]) hi=a[i];        }        scanf("%d",&k);        if(k==1)//特判排除了分母为0         {            printf("%lld\n",hi);            continue;        }        int ans;        while(lo<=hi)        {            mid=(lo+hi)>>1;            if(check(mid))            {                hi=mid-1;                 ans=mid;            }            else lo=mid+1;        }        printf("%d\n",ans);    }    return 0;}