B. Help Chef Gerasim

lb5645284

B. Help Chef Gerasim
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
[size=1em]
In a far away kingdom young pages help to set the table for the King. As they are terribly mischievous, one needs to keep an eye on the control whether they have set everything correctly. This time the royal chef Gerasim had the impression that the pages have
played a prank again: they had poured the juice from one cup to another. Now Gerasim wants to check his hypothesis. The good thing is that chef Gerasim always pour the same number of milliliters of juice to all cups in the royal kitchen. Having thoroughly
measured the juice in each cup, Gerasim asked you to write a program that will determine from which cup juice was poured to which one; otherwise, the program should determine that this time the pages set the table diligently.
[size=1em]
To simplify your task we shall consider the cups to be bottomless so that the juice never overfills a cup and pours out, however much it can be. Besides, by some strange reason in a far away kingdom one can only pour to a cup or from one cup to another an integer
number of milliliters of juice.

Input
[size=1em]
The first line contains integer n — the number of cups on the royal table (1 ≤ n ≤ 1000).
Next n lines contain volumes of juice in each cup — non-negative integers, not exceeding 104.

Output
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If the pages didn't pour the juice, print "Exemplary pages." (without the quotes). If you can determine the volume of juice poured during exactly one juice pouring,
print "v ml. from cup #a to
cup #b." (without the quotes), where v represents
the volume of poured juice, a represents the number of the cup from which the juice was poured (the cups are numbered with consecutive positive integers
starting from one in the order in which the cups are described in the input data), b represents the number of the cup into which the juice was poured. Finally,
if the given juice's volumes cannot be obtained using no more than one pouring (for example, the pages poured the juice from one cup to another more than once or the royal kitchen maids poured the juice into the cups incorrectly), print "Unrecoverable
configuration." (without the quotes).

Sample test(s)
input
5
270
250
250
230
250


output
20 ml. from cup #4 to cup #1.


input
5
250
250
250
250
250


output
Exemplary pages.


input
5
270
250
249
230
250


output
Unrecoverable configuration.

水题一道，题目大概意思是有一个侍从很顽皮，他把1-n个杯子果汁其中一杯倒进另外一杯中叫你求出是哪个杯子倒入哪个杯子并且倒了多少，不过如果侍从倒了多次或者果汁不能平均分就输出Unrecoverable configuration.如果侍从没有顽皮即每个杯子的柜子一样多久输出Exemplary
pages.
AC代码：
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<iomanip>
#include<cmath>
const int MAX=1001;
using namespace std;
typedef struct Cup
{
int v;
int num;
}cup;
cup s[MAX];
bool cmp(cup a,cup b)
{
return a.v<b.v;
}
int main()
{
int n,i,j,aver,count,p;
double sum;
cin>>n;
sum=0;
for(i=1;i<=n;i++)
{
cin>>s.v;
s.num=i;
sum+=double(s.v);
}
aver=int(sum/n);
if(aver==sum/n)
{
count=0;
for(i=1;i<=n;i++)
{
if(s.v==aver)
count+=1;
}
if(count==n)
cout<<&quot;Exemplary pages.&quot;<<endl;
else
{
sort(s+1,s+n+1,cmp);
p=0;
for(i=1;i<=n/2;i++)
{
if(((s.v+s[n-i+1].v)/2==aver)&&(s.v!=aver&&s[n-i+1].v!=aver))
{
p+=1;
}
}
if(p==1)
{
for(i=1;i<=n/2;i++)
{
if(((s.v+s[n-i+1].v)/2==aver)&&(s.v!=aver&&s[n-i+1].v!=aver))
{
cout<<(s[n-i+1].v-s.v)/2<<&quot; ml. from cup #&quot;<<s.num<<&quot; to cup #&quot;<<s[n-i+1].num<<'.'<<endl;
break;
}
}
}
else
cout<<&quot;Unrecoverable configuration.&quot;<<endl;
}
}
else
cout<<&quot;Unrecoverable configuration.&quot;<<endl;
}