Run ID | Author | Problem | Lang | Verdict | Score | Time | Memory | Code Length | Submit Time |
---|---|---|---|---|---|---|---|---|---|
6547 | 李言 | 排队打水 | C++ | Wrong Answer | 0 | 0 MS | 256 KB | 567 | 2020-10-02 11:16:43 |
#include<bits/stdc++.h> using namespace std; int main(){ int i,n,k=0,n1[101]={0},n2[101]={0}; double time=0; cin>>n; for(i=1;i<=n;i++){ cin>>n1[i]; n2[i]=n1[i]; } for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ if(n1[j]>n1[j+1]){ int t=n1[j]; n1[j]=n1[j+1]; n1[j+1]=t; } } } for(int i=1;i<=n+1;i++){ for(int j=1;j<=n;j++){ if(n1[i]==n2[j]){ cout<<j<<' '; n2[j]=0; break; } } } cout<<endl; for(int i=1;i<=n;i++){ time+=(n-i+1)*n1[i]; } printf("%.2lf",time/n); return 0; }