题意:
每件衣服都有一定单位水分,在不使用烘干器的情况下,每件衣服每分钟自然流失1个单位水分,但如果使用了烘干机则每分钟流失K个单位水分,但是遗憾是只有1台烘干机,每台烘干机同时只能烘干1件衣服,请问要想烘干N件衣服最少需要多长时间?
题解:待更新 ~~
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<algorithm>
#define ll long long
#define int ll
#define inf 0x3f3f3f3f
#define PI acos(-1.0)
using namespace std;
int read() {
int w = 1, s = 0;
char ch = getchar();
while (ch < '0' || ch>'9') {
if (ch == '-') w = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') {
s = s * 10 + ch - '0'; ch = getchar(); }
return s * w;
}
int gcd(int x, int y) {
return y ? gcd(y, x % y) : x; }
int ksm(int a, int b, int mod) {
int s = 1; while(b) {
if(b&1) s=s*a%mod;a=a*a%mod;b>>=1;}return s;}
const int N = 100010;
int n, k; //n件衣服 ,k是一分钟能烘干的水量
int a[N];//水分
bool check(int mid) {
int sum = 0;
for (int i = 0; i < n; ++i) {
//如果自然风干的时间不够,那么就需要使用洗衣机
if (a[i] > mid) {
int t = ceil((a[i] - mid) * 1.0 / (k - 1));//公式推导
sum += t;
}
}
return sum > x;
}
signed main()
{
while (~scanf("%lld", &n)) {
int maxx = 0;
for (int i = 0; i < n; ++i) {
scanf("%lld", &a[i]);
//可以取一个最大值作为二分的上界
maxx = max(maxx, a[i]);
}
k = read();//k为一分钟能烘干的水量
if (k == 1) {
printf("%lld\n", maxx);
continue;
}
int l = 0, r = maxx + 10;
while (l < r) {
int mid = l + r >> 1;
if (check(mid))
l = mid + 1;
else
r = mid;
}
printf("%lld\n", r);
}
return 0;
}