CF1132D Stressful Training

简介: CF1132D Stressful Training

题目: CF1132D Stressful Training ,哈哈,我们今天来看一道稍微复杂一点的题嘛,这是选自codeforce上的一道题,好了,我们一起来看看题意吧:

考虑到直接复制题目,或者截屏的方式不是很方便阅读,我就把直接题目链接放下面!

题目传送门: CF1132D Stressful Training

思路:

这道题思路就是采用贪心和二分的思想!!

我们先按照电脑能撑的时间时间从小到大排序,用一个优先队列来维护即可,我们每次判断队头是否符合条件即可,具体的直接看代码吧

我们来看看成功AC的代码吧:

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll n,k;
ll a[1000010],b[1000010];
ll l=0,r=2e12;
struct node{
    ll a,b;
    double c;
    bool operator < (const node &x)const {
        return c>x.c;
    }
};
priority_queue<node> q;
int check(ll x){
    while(!q.empty()) q.pop();
    for(int i=1;i<=n;i++){
        if(a[i]*1.0/b[i]<k){
            q.push({a[i],b[i],a[i]*1.0/b[i]});
        }
    }
    if(q.empty()) return 1; //若队列为空,说明当前x 取大了,因为 所有电脑都能在k分钟内不关机
    for(int i=0;i<k;i++){
        node t = q.top();   q.pop();
        t.a+=x;
        if(t.c<i) return 0;//这台电脑已经没电了
        t.c=t.a*1.0/t.b;
        if(t.a*1.0/t.b<k){
            q.push(t);
        }
        if(q.empty()) return 1;
    }
    return 1;
}
int main(){
    cin>>n>>k;
    for(int i=1;i<=n;i++) cin>>a[i];
    for(int i=1;i<=n;i++) cin>>b[i];
    while(l<r){
        ll mid=l+r>>1;
        if(check(mid)) r=mid;
        else l=mid+1;
    }
    if(l==2e12)
        cout<<-1<<"\n";
    else cout<<l<<"\n";
    return 0;
}


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