Description Title
has n individual line connected to a tap water before, if the water receiving each time is Ti, please find this programming individual queue n in a sequential, such that n individual minimum average waiting time.
Input and output format
input format:
Input common two rows, the first row n; second line respectively denote an n-th individual to individual contact of water per time T1, T2, ..., Tn, there is a space between each data.
Output formats:
There are two output lines, a first act of the queue sequence, i.e., an arrangement 1 to n; (accurate output result after two decimal places) the average waiting time in the second embodiment are arranged such behavior.
Input Output Sample
Input Sample # 1:
10
56 12 1 99 1000 234 33 55 99 812
Output Sample # 1:
32781496105
291.90
----------------
ideas: to traverse the entire array at the other person seeking person kick the accumulated total time spent on ans, Finally ans divided by n.
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
struct water{
int num;
int time;
}p[10010];
double ans;
inline bool cmp(water x,water y){
if(x.time!=y.time) return x.time < y.time;
return x.num < y.num;
}
int main(){
int n;
scanf("%d",&n);
for(int i = 1;i <= n;i++){
scanf("%d",&p[i].time);
p[i].num = i;
}
sort(p + 1,p + n + 1,cmp);
for(int i = 1;i <= n;i++)
printf("%d ",p[i].num);
printf("\n");
for(int i = 1; i<=n;i++)
ans += i*p[n-i].time;
ans /= n;
printf("%.2lf",ans);
return 0;
}
