Poj 3104 (二分)

Drying

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7104		Accepted: 1823

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 byk 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 containsai separated by spaces (1 ≤
ai ≤ 109). The third line containsk (1 ≤
k ≤ 109).

Output

Output a single integer — the minimal possible number of minutes required to dry all clothes.

Sample Input

sample input #132 3 95sample input #232 3 65

Sample Output

sample output #13sample output #22

Source

Northeastern Europe 2005, Northern Subregion

貪心的去類比會逾時，這裡關鍵是要想到二分。然後注意幾個細節。1 注意可能超int。 2在每次烘乾時，自然風乾是沒有的。 3 當k為1的情況。 4一個小技巧，整數運算中，a/b的上取整可以用（a+b-1）/b。

#include<cstdio>#include<algorithm>#include<vector>#include<set>#include<map>#include<iostream>#include<string>#include<cstring>#include<queue>#include<cmath>using namespace std;typedef long long LL;const int maxn = 100000 + 5;const int INF = 1000000000;int a[maxn];int n,k;bool can(int x){    int cnt = 0;    for(int i = 0;i < n;i++){        if(a[i] > x){            cnt += (a[i] - x + k - 2) / (k - 1);            if(cnt > x) return false;        }    }    return true;}int main(){    while(scanf("%d",&n) != EOF){        for(int i = 0;i < n;i++) scanf("%d",&a[i]);        sort(a,a+n);        scanf("%d",&k);        if(k == 1){            printf("%d\n",a[n-1]);            continue;        }        int l = 1,r = a[n-1];        int ans;        while(l <= r){            int mid = l+(r-l)/2;            if(can(mid)){                ans = mid;                r = mid-1;            }            else l = mid+1;        }        printf("%d\n",ans);    }    return 0;}