POJ 3104 Drying (二分&最大化最小值)

Posted 2022-12-07 synapse7

http://poj.org/problem?id=3104

一开始没注意k为1的情况，WA了一发。。

加个特判就过了。

完整代码：

/*750ms,576KB*/

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int mx = 100005;

int n, k, a[mx];

bool judge(int time)

	int cnt = 0;
	for (int i = 0; i < n; ++i)
	
		if (a[i] > time) cnt += ceil((double)(a[i] - time) / k);
		if (cnt > time) return false;
	
	return true;


int main()

	scanf("%d", &n);
	int l = 0, r = 0, m;
	for (int i = 0; i < n; ++i)  scanf("%d", &a[i]), r = max(r, a[i]);
	scanf("%d", &k);
	if (k == 1) return printf("%d", r), 0; ///要特判k为1的情况！！
	--k;
	while (l + 1 < r) judge(m = (l + r) >> 1) ? r = m : l = m;
	printf("%d", l + 1);
	return 0;