POJ 3104 二分

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 kwater, 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

题意：有n件湿度为wi （1<=wi<=10^9）的衣服，n件衣服各含ai水分，自然干一分钟一单位，放烘干机一分钟k单位，一次只能晒一件，用最少时间烘干这些衣服。

分析：

设需要用x分钟的机器，那么自然风干需要mid – x分钟，x和mid需要满足：

k*x + (mid – x) >= ai，即 x >= (ai – mid) / (k – 1)。当k=1的时候，很显然会发生除零错误，需要特殊处理

x是整数，要向上取整，用ceil.

函数名: ceil

用 法: double ceil(double x);

功 能: 返回大于或者等于指定表达式的最小整数

头文件:math.h

代码如下：

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
#include<cmath>
const int maxn=100005;
#define LL long long
int a[maxn],n,k;
bool C(int t)
{
    LL sum=0;
    for(int i=0;i<n;i++)
    {
        if(a[i]>t)
        {
            sum+=(int)ceil((double)(a[i]-t)/(k-1));
        }
    }
    return sum<=t;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int MAX=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            MAX=max(MAX,a[i]);
        }
        scanf("%d",&k);
        if(k==1)
        {
            printf("%d
",MAX);
            continue;
        }
        int lb=0,ub=MAX;
        while(ub-lb>1)
        {
            int mid=(lb+ub)>>1;
            if(C(mid))
            {
                ub=mid;
            }
            else
                lb=mid;
        }
        printf("%d
",ub);
    }
    return 0;
}