#include <iostream>
#include <algorithm>
#include<iomanip>
using namespace std; 
struct node{
	int id;
	int t;
}a[100005];
int n;
double sum = 0;
bool cmp(node a,node b)
{
  	if(a.t == b.t) return a.id < b.id;
	return a.t < b.t;
}
int main()
{
	cin >> n;
	for(int i = 0; i < n; i++)
	{
		cin >> a[i].t;
		a[i].id = i + 1;
	}
	sort(a, a + n, cmp);
	for(int i=0;i<n;i++)
  	{
  	    cout<<a[i].id<<" ";
  	    sum+=a[i].t*(n-i-1);
  	}
	cout << endl;
	cout << fixed << setprecision(2) <<sum/n;
	return 0;
}

排队接水

楼下dalao的解法稍微复杂了点，不用waitingtime这个数组也能解决。

思路很清晰：由于若把时间长的放在后面接水，那么就较少人等，所以排序+贪心即可。

对于每一个人，在场剩余每个人都要经历一次他的打水时间。

故遍历整个数组求其他人在这个人打水时花费的总时间累加到ans上，

最后ans除以n。 还要解决两位小数的保留问题！

提供两种方法

1.cout<<fixed<<setprecision(2)<<(头文件#include); 2.printf("%.2f",a)（这东西的头文件就不讲了）

上蒟蒻AC的代码

#include<queue>
#include<math.h>
#include<stdio.h>
#include<iostream>
#include<vector>
#include<iomanip>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<utility>
#include<cstring>
#include<stack>
#include<fstream>
#include<string>
using namespace std;
#define LL long long
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int n , s[1010];
long long sum;
struct lp
{
    int id;
    int m;
}a[1010];
int cmp( lp x ,lp y )
{
    if ( x.m < y.m )
    {
        return 1;
    }
    if ( x.m == y.m && x.id < y.id )
    {
        return 1;
    }
    return 0;
}
int main()
{
    cin >> n;
    for ( int i = 1 ; i <= n ; i++ )
    {
        cin >> a[i].m;
        s[i] = s[ i - 1 ] + a[i].m;
        a[i].id = i;
    }
    sort( a + 1 , a + n + 1 , cmp );
    for ( int i = 1 ; i <= n ; i++ )
    {
        cout << a[i].id << " ";
    }
    for ( int i = 1 ; i <= n ; i++ )
    {
        sum += a[i].m * ( n - i );
    }
    cout << fixed << setprecision(2) << endl << sum * 1.0 / n;
    return 0;
}

提交制作不易，麻烦点赞在离开😁😁😁

#include<bits/stdc++.h>
using namespace std; 
const int N=1e5+10;
    int i,n,sum;
    struct shui{
    	int id;
    	int su;
	}a[N];
	double ans;
int cmp(shui a,shui b){
	if(a.su==b.su)
	    return a.id<b.id;
	return a.su<b.su;
}
int main(){
	cin>>n;
    for(int i=1;i<=n;i++){
    	cin>>a[i].su;
    	a[i].id=i;
	}
	sort(a+1,a+n+1,cmp);
	for(int i=1;i<=n;i++){
		cout<<a[i].id<<" ";
		ans+=a[i].su+a[i].su*(n-i-1);
	}
    cout<<endl<<fixed<<setprecision(2)<<ans/n;
    return 0;
}

#include <iostream>
#include <algorithm>
#include<iomanip>
using namespace std; 
struct node{
	int id;
	int t;
}a[100005];
int n;
double sum = 0;
bool cmp(node a,node b)
{
  	if(a.t == b.t) return a.id < b.id;
	return a.t < b.t;
}
int main()
{
	cin >> n;
	for(int i = 0; i < n; i++)
	{
		cin >> a[i].t;
		a[i].id = i + 1;
	}
	sort(a, a + n, cmp);
	for(int i=0;i<n;i++)
  	{
  	    cout<<a[i].id<<" ";
  	    sum+=a[i].t*(n-i-1);
  	}
	cout << endl;
	cout << fixed << setprecision(2) <<sum/n;
	return 0;
}

#include <iostream>
#include <bits/stdc++.h> 
using namespace std;
int n,s[1005];long sum;
struct lp{
	int id;
	int m;
}a[1005];
int cmp(lp x,lp y)
{
	if(x.m<y.m){
		return 1;
	}
	if(x.m==y.m&&x.id<y.id)return 1;
	return 0;
}
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i].m;
		s[i]=s[i-1]+a[i].m;
		a[i].id=i;
	}
	sort(a+1,a+n+1,cmp);
	for(int i=1;i<=n;i++)
	{
		cout<<a[i].id<<" ";
	}
	for(int i=1;i<=n;i++)
	{
		sum+=a[i].m*(n-i);
	}
	cout<<fixed<<setprecision(2)<<endl<<sum*1.0/n;
	return 0;
}

#include <bits/stdc++.h> 
using namespace std;
const int N=1e5+10;
struct stu{
	int id,t;
}a[1005];
int cmp(stu a,stu b){
	if(a.t<b.t) return 1;
	else return 0;
}
int main(){
	int n;
	cin>>n;
	sort(a,a+n,cmp);
	for(int i=1;i<=n;i++){
		cin>>a[i].t;
		a[i].id=i;
	}
	sort(a+1,a+n+1,cmp);
	for(int i=1;i<=n;i++)
		cout<<a[i].id<<" ";
	cout<<endl;
	double ans=0,time;
	for(int i=1;i<n;i++){
		time=a[i].t*(n-i);
		ans+=time;
	}
	ans/=n;
	cout<<fixed<<setprecision(2)<<ans;
}

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<algorithm>
#define LL long long
using namespace std;
const int N=1e5+10;
const int Inf=0x3f3f3f3f;
int n;
double time1 = 1.00 * 0;
struct per {
	int t;
	int num;
}a[N];

bool cmp(per a, per b){
	if(a.t==b.t){
		return a.num<b.num;
	}
	return a.t < b.t;
}

int main(){
	int n;
	cin >> n;
	for(int i = 1; i <= n; i++) {
		cin >> a[i].t;
		a[i].num = i;
	}
	sort(a + 1, a + n + 1, cmp);
	int sum = n - 1;
	for(int i = 1; i <= n; i++) {
		time1 += 1.00 * a[i].t * sum;
		sum--;
		cout << a[i].num << ' ' ;
	}
	time1 /= 1.00 * n;
	cout << endl;
	printf("%.2lf", time1);
}

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+10;
const int INF=0x3f3f3f3f;
int n;
double tim = 1.00 * 0;
struct person {
	int t;
	int num;
}a[N];

bool cmp(person a, person b)
{
	return a.t < b.t;
}

int main()
{
	int n;
	cin >> n;
	for(int i = 1; i <= n; i++) {
		cin >> a[i].t;
		a[i].num = i;
	}
	sort(a + 1, a + n + 1, cmp);
	int sum = n - 1;
	for(int i = 1; i <= n; i++) {
		tim += 1.00 * a[i].t * sum;
		sum--;
		cout << a[i].num << ' ' ;
	}
	tim /= 1.00 * n;
	cout << endl;
	printf("%.2lf", tim);
}
40分