1 solutions

  • 1
    @ 2024-3-31 9:13:46
    #include<bits/stdc++.h>
    using namespace std;
    const int N=50010;
    int a[N];
    int L,n,m;
    bool check(int mid)
    {
        int last=0,cnt=0; //上一个石头的位置和移走的石头数目 
        for(int i=1;i<=n;i++)
        {
            if(a[i]-last>=mid) last=a[i]; //更新上一个石头的距离 
            else cnt++; //移走 
        }
        return cnt<=m; //说明距离可以更大 
    }
    int main()
    {
        cin>>L>>n>>m;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        n++;
        a[n]=L;
        int l=0,r=1e9;
        while(l<r) //枚举当前跳跃的最短距离
        {
            int mid=l+r+1>>1;
            if(check(mid)) l=mid; //当前长度可以 
            else r=mid-1;
        }
        cout<<l;
        return 0;
    }
    
    • 1

    Information

    ID
    953
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    5
    Tags
    (None)
    # Submissions
    42
    Accepted
    18
    Uploaded By