poj 3104 最大化最小值

链接:https://cn.vjudge.net/problem/POJ-3104

因为每分钟烘干K只比普通风干少k-1，所以是除k-1

#include 
#include 
#include 
#include 
using namespace std;
 
const int N = 100010;
typedef long long LL;
LL num[N];

int main(){
	int n;
	while(scanf("%d",&n) != EOF){
		LL maxvalue = 0,rp = 0;
		memset(num,0,sizeof(num));
		for(int i = 0; i < n; ++i){
		   scanf("%lld",&num[i]);
		   if(num[i] > maxvalue)
			   maxvalue = num[i];
		}
		LL k;
		scanf("%lld",&k);
		if(k == 1){
		  printf("%lld\n",maxvalue);
		  continue;
		}
		LL lp = 1,ans = 0;
		rp = maxvalue;
		while(lp <= rp){
		   LL mid = (lp + rp) / 2, sum  = 0;
		   for(int i = 0; i < n; ++i){
			   if(num[i] > mid){
			     long long xx = ceil((num[i] - mid ) * 1.0 / (k - 1) );
				 sum += xx;
			   }
		   }
		   //cout<