POJ 2456 Aggressive cows(二分答案，最大化最小值)

#include<cstdio>
#include<algorithm>

//#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define X first
#define Y second
#define cl(a,b) memset(a,b,sizeof(a))
typedef pair<long long ,long long > P;
const int maxn=100005;
const LL inf=1LL<<45;
const LL mod=1e9+7;

LL a[maxn];
int n,m;
bool ok(int d){
int last=0;
for(int i=1;i<m;i++){
int cur=last+1;
while(cur<n&&a[last]+d>=a[cur])cur++;
if(cur==n)return false;
last=cur;
}
return true;
}
void solve(){
int l=0,r=999999999;
while(r-l>1){
int mid=(l+r)/2;
if(ok(mid))l=mid;
else r=mid;
}
printf("%d\n",r);
}
int main(){
while(~scanf("%d%d",&n,&m)){
for(int i=0;i<n;i++){
scanf("%lld",&a[i]);
}
sort(a,a+n);
solve();
}
return 0;
}