Codeforces Round #417 (Div. 2) C. Sagheer and Nubian Market

C. Sagheer and Nubian Market

传送门

二分k，首先O(n)得到改变后的数组，用优先队列得到前k各最小值，总的时间复杂度是O(nlogn)

#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <vector>
#define ll unsigned long long
#define inf 1000000000LL
#define mod 1000000007
using namespace std;
int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
const int N=1e5+10;
int a[N],n;
ll sum[N];
ll check(ll k){
   priority_queue<ll,vector<ll>,greater<ll> >que;
   for(int i=1;i<=n;i++){
       ll x=a[i]+i*k;
       que.push(x);
   }
   ll ans=0;
   for(int i=1;i<=k;i++){
      ans+=que.top();
      que.pop();
   }
   return ans;
}
int main()
{
   // while(true)
    {
        n=read();
        int m=read();
        for(int i=1;i<=n;i++) a[i]=read();
        int l=0,r=n;
        ll ans,cans;
        ll sd=m;
        while(l<=r){
             ll mid=(l+r)>>1;
             ll tmp=check(mid);
             if(tmp<=sd){
                 l=mid+1;
                 ans=mid;
                 cans=tmp;
             }
             else r=mid-1;
        }
        printf("%I64d %I64d
",ans,cans);
    }
    return 0;
}