poj3258 二分（连续区间求和）

#include<iostream>#include<algorithm>using namespace std;#define maxn 50010long long a[maxn],L,b[maxn];int m,n;bool ok(long long x){    int flag=0,pre=0;    long long sum=0;    for(int j=1;j<=n+1;j++)    {        if(a[j]-a[pre]<x)//小于x的点需要去掉        {            flag++;        }        else   //大于x的点直接跳过        {            pre=j;        }    }    if(flag<=m){ return true;}    else return false;}int main(){    int i;    cin>>L>>n>>m;    a[0]=0;    a[n+1]=L;    for(i=1;i<=n;i++)    {        cin>>a[i];    }    sort(a,a+n+2);    long long low=0,high=L,mid,ans=0;    while(low<=high)    {        mid=(low+high)/2;        if(ok(mid))        {            ans=mid,low=mid+1;        }        else high=mid-1;    }    cout<<ans<<endl;}