Existing n pieces of clothes need to be dried, and the moisture content of each piece of clothing is ai. If it is dried naturally, the moisture content will decrease by 1 per minute; if the dryer is used for drying, the moisture content will decrease by k (until 0). There is only one dryer, which can only dry one piece of clothing at a time, and use it for at least 1 minute at a time. What is the minimum time to make all the clothes water content 0?
#include<stdio.h>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN = 1e5+10;
int water[MAXN];
//判断在时间time内是否可以将所有衣服烘干
bool judge(int n,int k,int time){
int sum = 0;
for(int i = 0;i < n; ++i){
if(water[i] > time){
sum += ceil((water[i] - time)*1.0 / (k-1));
}
if(sum > time)
return false;
}
return true;
}
int main()
{
int n;
scanf("%d",&n);
for(int i = 0;i < n; ++i){
scanf("%d",&water[i]);
}
int k;
scanf("%d",&k);
sort(water,water+n);
if(k == 1)
printf("%d\n",water[n-1]);
else{
//二分法找满足条件的最小值
int left = 1;
int right = water[n-1];
while(left <= right){
int middle = left + (right - left)/2;
if(judge(n,k,middle)){
right = middle - 1;
}
else{
left = middle + 1;
}
}
printf("%d",left);
}
return 0;
}