【POJ 3104】Drying

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
2

题目大意：

　　有N件衣服需要晒，现在有一台烘干机，烘一分钟相当于晒K分钟，但是只能同时烘干一件衣服，现在给出每件衣服需要晒的时间，求出晒干所有衣服所需要的最少时间。

分析：

　　二分答案，判断可行性。

　　假如二分到的答案为mid，那么需要晒的时间小于等于mid的就都拿去晒。

　　大于t的，假设该衣服晒a分钟，烘干b分钟，那么需要满足：a + b <= mid, a + bk >= t。

　　可以转换成b >= (t - mid) / (k - 1)，因为y是整数，所以b最小是b = (t - mid + k - 2) / (k - 1)

　　把所有b加起来，小于等于mid则可行。

　　PS：judge ()里面的sum不开long long会爆QAQ。

代码：

#include <cstdio>

int n, k, i, t[100010];
int left, right, mid;

bool judge (int lim)
{
    long long sum = 0;
    for (i = 0; i < n; i++)
        if (t[i] > lim) sum += (t[i] - lim + k - 2) / (k - 1);
    return sum <= lim;
}

int main ()
{
    right = !scanf ("%d", &n);
    for (i = 0; i < n; i++)
    {
        scanf ("%d", &t[i]);
        if (t[i] > right) right = t[i];
    }
    scanf ("%d", &k);
    if (k > 1)
    {
        for (left = 0; left < right; )
        {
            mid = left + right >> 1;
            if (judge (mid)) right = mid;
            else left = mid + 1;
        }
    }
    printf ("%d", right);
}