POJ 3104 Drying (二分答案)

题目链接：http://poj.org/problem?id=3104

Drying

Time Limit: 2000MS

Memory Limit: 65536K
Total Submissions: 18125

Accepted: 4570

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

题解：

  每分钟都能风干一个水分，烘干机每分钟能烘干K个水分，可以假想K个水分中，一个是风干的，（K-1）个是烘干机烘干的。
  题目满足二分条件，小的时间如果可以，那么更大时间一定也可以，那么二分答案，验证答案的方法，已知烘干时间为time，那么所有衣服都会自然风干time个水分，其他的烘干机来烘干，因为必须烘干到0个水分，所以用（a[i]-time）/(k-1) 向上取整求得。
  需要特殊处理k等于1的时候。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#define mid ((l+r)>>1)using namespace std;const int N = 100005;int n, k;int a[N];bool check(int time) {    int ans = 0;    for(int i = 0; i < n; i++)    if(a[i]-time > 0) {        if(k == 1) return 0;  // 特判k==1        ans += ceil((a[i]-time)*1./(k-1));        if(ans > time) return 0;    }    return true;}int main() {    while(~scanf("%d", &n)) {        int l = 1;        int r = 0;        for(int i = 0; i < n; i++) {            scanf("%d", &a[i]);            r = max(r, a[i]);        }        scanf("%d", &k);        while(l < r) {            if(succ(mid))                r = mid;            else                l = mid+1;        }        printf("%d\n", r);    }    return 0;}