练习cf1490B. Balanced Remainders

题目如下
B. Balanced Remainders
time limit per test2 seconds
memory limit per test256 megabytes
You are given a number 𝑛 (divisible by 3) and an array 𝑎[1…𝑛]. In one move, you can increase any of the array elements by one. Formally, you choose the index 𝑖 (1≤𝑖≤𝑛) and replace 𝑎𝑖 with 𝑎𝑖+1. You can choose the same index 𝑖 multiple times for different moves.

Let's denote by 𝑐0, 𝑐1 and 𝑐2 the number of numbers from the array 𝑎 that have remainders 0, 1 and 2 when divided by the number 3, respectively. Let's say that the array 𝑎 has balanced remainders if 𝑐0, 𝑐1 and 𝑐2 are equal.

For example, if 𝑛=6 and 𝑎=[0,2,5,5,4,8], then the following sequence of moves is possible:

initially 𝑐0=1, 𝑐1=1 and 𝑐2=4, these values are not equal to each other. Let's increase 𝑎3, now the array 𝑎=[0,2,6,5,4,8];
𝑐0=2, 𝑐1=1 and 𝑐2=3, these values are not equal. Let's increase 𝑎6, now the array 𝑎=[0,2,6,5,4,9];
𝑐0=3, 𝑐1=1 and 𝑐2=2, these values are not equal. Let's increase 𝑎1, now the array 𝑎=[1,2,6,5,4,9];
𝑐0=2, 𝑐1=2 and 𝑐2=2, these values are equal to each other, which means that the array 𝑎 has balanced remainders.
Find the minimum number of moves needed to make the array 𝑎 have balanced remainders.

Input
The first line contains one integer 𝑡 (1≤𝑡≤104). Then 𝑡 test cases follow.

The first line of each test case contains one integer 𝑛 (3≤𝑛≤3⋅104) — the length of the array 𝑎. It is guaranteed that the number 𝑛 is divisible by 3.

The next line contains 𝑛 integers 𝑎1,𝑎2,…,𝑎𝑛 (0≤𝑎𝑖≤100).

It is guaranteed that the sum of 𝑛 over all test cases does not exceed 150000.

Output
For each test case, output one integer — the minimum number of moves that must be made for the 𝑎 array to make it have balanced remainders.

题目大意
现有一个有n个元素的数组，n为3的除数（可以被3整除），每次可以对一个元素进行加一或者减一的操作，问最少需要多少次操作，能够使得数组中整除3得1，整除3得2和整除3得0的元素数量相等。

题目分析
先分组，统计c1，c2，c0分别的个数，再通过搬运获得相同的数。

点击查看代码

#include <iostream>
using namespace std;

int main(){
    int t;
    cin >> t;
    while(t--){
        int n;
        cin >> n;
        int a[300003];
        int c1 = 0, c2 = 0, c0 = 0, cnt = 0;
        for(int i = 0; i < n; i++){
            cin >> a[i];
            if(a[i] % 3 == 0){
                c0++;
            }else if(a[i] % 3 == 1){
                c1++;
            }else{
                c2++;
            }
        }
        while(c0 != n / 3 || c1 != n / 3 || c2 != n / 3){ //
            if(c0 > n / 3){
                c0--;
                c1++;
                cnt++;
            }else if(c1 > n / 3){
                c1--;
                c2++;
                cnt++;
            }else if(c2 > n / 3){
                c2--;
                c0++;
                cnt++;
            }
        }
        cout << cnt<< endl;
    }
}