AC Saber二分

Posted 2021-08-21 辉小歌

目录

• 数的范围
• 数的三次方根
• 最佳牛围栏
• 特殊排序

数的范围

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=1e5+10;
int a[N],n,m,x;
int main(void)
{
    cin>>n>>m;
    for(int i=0;i<n;i++) cin>>a[i];
    while(m--)
    {
        cin>>x;
        int l=0,r=n-1;
        while(l<r)
        {
            int mid=l+r>>1;
            if(a[mid]>=x) r=mid;
            else l=mid+1;
        }
        if(a[l]==x) cout<<l<<" ";
        else cout<<-1<<" ";
        l=0,r=n-1;
        while(l<r)
        {
            int mid=l+r+1>>1;
            if(a[mid]<=x) l=mid;
            else r=mid-1;
        }
        if(a[l]==x) cout<<l<<endl;
        else cout<<-1<<endl;
    }
    return 0;
}

数的三次方根

#include<cstdio>
#include<iostream>
using namespace std;
double n; 
bool check(double x)
{
    if(x*x*x<=n) return true;
    else return false;
}
int main(void)
{
    cin>>n;
    double l=-100,r=100;
    while(r-l>=1e-8)
    {
        double mid=(l+r)/2;
        if(check(mid)) l=mid;
        else r=mid;
    }
    printf("%.6lf\\n",l);
    return 0;
}

最佳牛围栏

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=1e5+10;
int a[N];
double sum[N],n,m;
bool check(double avg)
{
    for(int i=1;i<=n;i++) sum[i]=sum[i-1]+a[i]-avg;
    
    double mina=99999;
    for(int i=0,j=m;j<=n;j++,i++)
    {
        mina=min(mina,sum[i]);
        if(sum[j]>=mina) return true;
    }
    return false;
}
int main(void)
{
    cin>>n>>m;
    for(int i=1;i<=n;i++) cin>>a[i];
    double l=1,r=2000;
    while(r-l>1e-7)
    {
        double mid=(l+r)/2;
        if(check(mid)) l=mid;
        else r=mid;
    }
    printf("%d\\n",int(r*1000));
    return 0;
}

特殊排序

// Forward declaration of compare API.
// bool compare(int a, int b);
// return bool means whether a is less than b.

class Solution {
public:
    vector<int> specialSort(int N) {
        vector<int> res; res.push_back(1);
        for(int i=2;i<=N;i++)
        {
            int l=0,r=res.size()-1;
            while(l<r)
            {
                int mid=l+r+1>>1;
                if(compare(res[mid],i)) l=mid;
                else r=mid-1;
            }
            res.push_back(i);
            for(int j=res.size()-2;j>r;j--) swap(res[j],res[j+1]);
            if(compare(i,res[r])) swap(res[r],res[r+1]);
        }
        return res;
    }
};