Drying（二分搜索）

Drying

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8126		Accepted: 2064

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

 

       题意：

       给出 N （1 ~ 100000），代表有 N 件衣服，后给出每件衣服的湿度 ai（1 ~ 10 ^ 9），后给出 K （1 ~ 10 ^ 9）代表烘干机的烘干速度是 K，平常风干的话是 1。烘干机每次只能对一件衣服，输出当所有衣服都干了的最少时间。

 

       思路： 

       二分搜索。搜索烘干时间 t 。找最小值，所以左开右闭，t 的范围是 （0 ，max_ai ] 。假设烘干的时间为 t ，那么就要求所有衣服都必须要在 t 时间内变干：

       若 ai <= t ， 直接风干就好，不需要烘干机；

       若 ai > t ，则需要烘干机，因为每次烘干机只对一件衣服，所以要最大利用烘干机的次数才行，对于每件衣服来说，就需要求一个最少利用烘干机的时间数。

       设风干的时间为 F，烘干机的时间为 H，所以 F + H = t ；对于 ai，F + H * k >= ai，两条式子合并消掉 F ，就能得到 H >= ( ai - t ) / (k - 1) ，对得数向上取整便求出最少的 H 。将所有衣服的 H 相加起来可以得三种情况：

       若 H < t ，说明在规定的时间 t 内能把全部衣服给烘干，那么应该继续往小的搜，所以搜 ( l , t ]；

       若 H > t ，说明在规定的时间 t 内不能把全部衣服烘干，那么应该继续往大的搜，所以搜 ( t , r ]；

       若 H == t ，说明在规定的时间 t 内能把全部衣服烘干，但是要求找到的是最少的时间，所以要继续往小的搜，所以搜 (l , t ]。

 

       注意点：

       1.注意搜索边界，是从 0 到 max_ai，而不是 min_ai 到 max_ai；

       2.注意推到公式的条件，分母为k - 1，所以 k - 1 != 0，所以当 k == 1 的时候要单独拿出来讨论；

       3.注意公式的正确推到过程；

       4.二分搜索边界要开 long long 型，不然会WA；

 

       AC：

#include <cstdio>
#include <algorithm>
const int INF = 100000005;
using namespace std;

typedef long long ll;

int n, max_t, k;
int clo[100005];

bool C(ll t) {
        ll ans = 0;

        for (int i = 0; i < n; ++i) {
                if (clo[i] > t) {
                        ll left = clo[i] - t;
                        ans += (left / (k - 1) + (left % (k - 1) != 0));
                }
        }

        return ans <= t;
}

void solve() {
        ll l = 0, r = max_t;

        while(r - l > 1) {
                ll mid = l + (r - l) / 2;
                if (C(mid)) r = mid;
                else l = mid;
        }

        printf("%lld\n",r);
}

int main() {
        //freopen("test.in", "r", stdin);

        scanf("%d", &n);

        max_t = -1;
        for (int i = 0; i < n; ++i) {
                scanf("%d", &clo[i]);
                max_t = max(max_t,clo[i]);
        }

        scanf("%d", &k);

        if(k == 1) printf("%d\n",max_t);
        else solve();

        return 0;
}