POJ-3104

还是一道二分搜索，比较简单，针对时间二分，有一点需要特别注意，就是如果用风干机的话，每分钟减k点水，这时候就不会再减自然风干的那1点了，否则的话就每分钟减1点，不要算多了

#include<cstdio>
#include<algorithm>

using namespace std;

namespace
{
	int n, k, dat[100000], maxx;

	bool check(int time)
	{
		int count = time;
		for (int i = 0; i < n; i++)
		{
			if (dat[i] > time)
			{
				int remain = dat[i] - time;
				count -= remain % k ? remain / k + 1 : remain / k;
			}
			if (count < 0)
				return false;
		}
		return true;
	}

	void solve()
	{
		if (k == 1)
			printf("%d\n", maxx);
		else
		{
			k--;
			int lb = 0, ub = maxx, mid;
			while (lb + 1 < ub)
			{
				mid = (lb + ub) / 2;
				if (check(mid))
					ub = mid;
				else
					lb = mid;
			}
			printf("%d\n", ub);
		}
	}
}

int main()
{
	scanf("%d", &n);
	maxx = 0;
	for (int i = 0; i < n; i++)
	{
		scanf("%d", &dat[i]);
		maxx = max(maxx, dat[i]);
	}
	scanf("%d", &k);
	solve();
	return 0;
}