candy - candy i spoj solution
#include <iostream>
#include <cstdio>
using namespace std;
int main(void) {
long long i,j,k,n,sum,a[100001],less,more;
while(1)
{
scanf("%lld",&n);
//n==-1 indicates the end of the input or values
if(n==-1)
break;
//Initialize less , more and sum with "0"
less=0;
more=0;
sum=0;
for(i=1;i<=n;i++)
{
scanf("%lld",&a[i]);
}
for(i=1;i<=n;i++)
sum=sum+a[i];
sum=sum/n;
//sum here indicates number of candies that should be
//there in each packets from here after
for(i=1;i<=n;i++)
{
//calculating more and less as mentioned in our explanation
if(a[i]<sum)
less+=(sum-a[i]);
else
more+=(a[i]-sum);
}
//Incase less and more are not equal then we can't
// keep equal number of candies in all packets
//according to given input otherwise print the output
if(less!=more)
printf("-1\n");
else
printf("%lld\n",more);
}
return 0;
}
