POJ 3258 River Hopscotch（二分）

#pragma warning(disable:4996)
#include <cstdio>
#include <algorithm>
using namespace std;
int a[50005];
int l, n, m;

int get(int x){//返回距离最小值>=x时去掉的石头数量
int cnt = 0, i = 1, last = 0;
for (i = 1; i <= n + 1; i++){
if (a[i] - last < x)cnt++;
else last = a[i];
}
return cnt;
}

int main(){
scanf("%d %d %d", &l, &n, &m);

a[0] = 0;
for (int i = 1; i <= n; i++)scanf("%d", a + i);
a[n + 1] = l;
sort(a + 1, a + 1 + n);

int low = 0, high = l;
while (low + 1 < high){
int mid = (low + high) >> 1;
int cnt = get(mid);
if (cnt == m)low = mid;//若去掉的石头数量为m，那么答案>=mid，所以low=mid
else if (cnt > m)high = mid - 1;//若去掉的石头数量大于m，那么答案<mid，所以high=mid-1
else low = mid + 1;//若去掉石头数量小于m，那么答案>mid，所以low=mid+1
}

if (get(high) == m)printf("%d\n", high);
else printf("%d\n", low);

return 0;
}