POJ 3104

Drying

Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu

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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 #132 3 95sample input #232 3 65

Sample Output

sample output #13sample output #22

这题虽然说要用二分思想，可是在计算最少时间时还涉及到数学推导

题解：第一时间就想到使用二分枚据答案+验证这种思路，不过这题还是有些陷阱需要注意。1. 验证答案时，如果 a[i] <= mid，让它自然烘干即可 ；

 如果a[i] > mid，那么烘干这件衣服可以分成两段时间：（注意要使时间最短）使用烘干机时间x1 + 自然烘干时间x2，

那么可以列出等式：mid = x1 + x2;

 a[i] <= kx1+x2；于是得x1 >= (a[i] -mid)/(k-1);即得使用烘干机的最少时间x1

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int N = 100010;long long a[N];int judge(long long x);int n;long long k;int main(){    while(scanf("%d", &n)!=EOF)    {        long long  l=1, r=0;        for(int i=1;i<=n;i++)        {            scanf("%lld", &a[i]);            r=max(r,a[i]);        }        scanf("%lld", &k);        if(k==1)        {            printf("%lld\n",r);            continue;        }        while(l<=r)        {            long long mid=(l+r)/2;            if(judge(mid))            {                r=mid-1;            }            else            {                l=mid+1;            }        }        printf("%lld\n",l);    }    return 0;}int judge(long long x){    long long cnt=0;    for(int i=1;i<=n;i++)    {        if(a[i]>x)        {            cnt=cnt+ceil((a[i]-x)*1.0/(k-1));        }    }    if(cnt<=x)    {        return 1;    }    return 0;}

Drying

Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu

Submit Status

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 #132 3 95sample input #232 3 65

Sample Output

sample output #13sample output #22