Codeforces Round #702 (Div. 3)B. Balanced Remainders

思路：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int mod = 1e9 + 7;
int c[3];
int main()
{
  int t;
  scanf("%d",&t);
  while(t--)
  {
    memset(c,0,sizeof(c));
    int n;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
    {
      int a;
      scanf("%d",&a);
      c[a%3]++;
    }
    int pos,k = n/3;
    int ans = 0;
    while(c[0]!=c[1]||c[0]!=c[2]||c[1]!=c[2])
    {
      for(int i = 0; i < 2; i++)
      {
        if(c[i] >= c[i+1])
        {
          pos = i;
          break;
        }
      }
      if(c[2] > c[0] && c[2] > c[1])
      pos = 2;
      ans += (c[pos] - k);
      c[(pos+1)%3] += (c[pos] - k);
      c[pos] = k;
    }
    printf("%d\n",ans);
  }
  return 0;
}