Run ID | Author | Problem | Lang | Verdict | Time | Memory | Code Length | Submit Time |
---|---|---|---|---|---|---|---|---|
36144 | 王皓暄 | [贪心算法]排队接水 | C++ | Accepted | 0 MS | 244 KB | 524 | 2024-01-08 19:29:35 |
#include<bits/stdc++.h> using namespace std; struct s { int num,time; }; int main() { int n; cin>>n; s ns[n]; for(int i=0;i<n;i++) { cin>>ns[i].time; ns[i].num=i+1; } for(int i=0;i<n;i++) { int max=i; for(int j=i;j<n;j++) { if(ns[j].time<ns[max].time)max=j; } s t=ns[max]; ns[max]=ns[i]; ns[i]=t; } int sum=0; for(int i=0;i<n;i++) { cout<<ns[i].num<<' '; sum+=ns[i].time*(n-1-i); } cout<<endl; printf("%.2f",sum*1.0/n); return 0; }