POJ 3104 Drying Clothes

C - Drying Clothes

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3104

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 32 3 95 sample input #2 32 3 65

Sample Output

sample output #1 3 sample output #2 2

二分答案，先假设全部自然晾干，对于还余下的水分需要使用烘干机ceil(x*1.0/(k-1))次，看是否符合要求次数（时间）。

注意：分母为0要单独讨论。整形时会运行时错误，但实型不会报错，返回一个很大的数（inf）。

#include <iostream>#include<cmath>#include<cstdio>using namespace std;int a[100100],k,ma;int n;long long  check(long long t){  long long i,j,ans=0;  for (i=1;i<=n;i++){    long long x;    x=a[i]-t;    if (x>0) ans+=ceil(x*1.0/(k-1));  }  if (ans<=t) return 1;  else return 0;}long long bse(){  long long l,r,m,i,j;  l=1;r=ma;m=(l+r)/2;  while (l<r){    if (check(m)) r=m;    else l=m+1;    m=(l+r)/2;  }  return m;}int main(){    long long i;    scanf("%d",&n);    ma=0;    for (i=1;i<=n;i++) {        scanf("%d",&a[i]);        ma=max(ma,a[i]);    }    scanf("%d",&k);    if (k==1) printf("%d\n",ma);    else    printf("%d\n",bse());    return 0;}