POJ2456 Aggressive cows

#include <cstdio>
#include <algorithm>

using namespace std;
const int maxl = 100010;
int a[maxl];
int N, M;

inline bool condition(int d) 
{
	int pr = 0;
	for (int i = 1; i < M; ++i) {     //第一次循环可看作是找1 ，2两头牛的距离，第M － 1 次寻找M － 1 和第M头牛之间的距离
		int ne = pr + 1;
		while (ne < N && a[ne] - a[pr] < d) ++ne;
		if (ne == N) return false;    //要符合当前距离，但是牛舍“不够用了”，距离偏大
		pr = ne;
	}
	return true;
}
int main()
{
	while (scanf("%d%d", &N, &M) != EOF) {
		int maxv = 0;
		for (int i = 0; i < N; ++i) {
			scanf("%d", &a[i]);
			if (a[i] > maxv) maxv = a[i];
		}
		sort(a, a + N);
		//二分
		int lp = 0, rp = maxv;
		while (rp - lp > 1) {
			int mid = (lp + rp) / 2;
			if (condition(mid)) lp = mid;
			else rp = mid;
		}
		printf("%d\n", lp);
	}
	return 0;
}