poj 2456 Aggressive cows

题目：

首先把牛棚的位置x[i]按从小到大排序。牛棚数量N，牛数量为C，则每两只牛间可能的最大距离r=(x[N - 1] - x[0]) / (C - 1)。用二分法搜索0~r,判断mid满不满足条件。

AC情况：

代码：

#include<iostream>#include<algorithm>using namespace std;int x[100005];int N, C;int main() {    scanf("%d%d", &N, &C);    for (int i = 0; i < N; i++)        scanf("%d", &x[i]);    sort(x, x + N);    int l = 1, r = (x[N - 1] - x[0]) / (C - 1);    while (r - l > 1) {        int mid = (l + r) / 2, ans = 0;        for (int i = 0; ans < C&&i < N;) {            int t = i;            while (x[++i] - x[t] < mid);            ans++;        }        if (ans >= C)            l = mid;        else            r = mid;    }    if (l != r) {        int ans = 0;        for (int i = 0; ans < C&&i < N;) {            int t = i;            while (x[++i] - x[t] < r);            ans++;        }        if (ans >= C)            l = r;    }    printf("%d\n", l);    return 0;}