Aggressive cows - POJ 2456 - Virtual Judge

分析：

二分条件：某个点之后的区间具有什么特殊性质，能够将这个点二分出来。

枚举最小值，判断是否符合条件。

代码：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>using namespace std;const int N = 1e5 + 10;int n, c;
int x[N];inline int read()
{int sum = 0, ch = getchar();while (ch < '0' || ch > '9') ch = getchar();while (ch >= '0' && ch <= '9'){sum = sum * 10 + ch - '0';ch = getchar();}return sum;
}bool check(int k)
{int pos = x[0];int cnt = 1;for (int i = 1; i < n; i ++ ){if (x[i] - pos >= k) {pos = x[i];cnt ++ ;}if (cnt >= c) return true;}return false;
}int main()
{scanf("%d%d", &n, &c);for (int i = 0; i < n; i ++ ) scanf("%d", &x[i]);sort(x, x + n);//左边界从0开始！！！int l = 0, r = x[n - 1] - x[0];int ans = 0;while (l < r){int mid = l + r + 1 >> 1;if (check(mid)) l = mid, ans = mid;else r = mid - 1;}printf("%d\n", ans);return 0;
}