CoderForces999D-Equalize the Remainders

D. Equalize the Remainders

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array consisting of $n$ integers $a_1,a_2,\dots ,a_n$, and a positive integer $m$. It is guaranteed that $m$ is a divisor of $n$.

In a single move, you can choose any position $i$ between $1$ and $n$ and increase $a_i$ by $1$.

Let's calculate $c_r$ ($0\le r\le m-1)$ — the number of elements having remainder $r$ when divided by $m$. In other words, for each remainder, let's find the number of corresponding elements in $a$ with that remainder.

Your task is to change the array in such a way that $c_0=c_1=\cdots =c_{m-1}=\frac{n}{m}$.

Find the minimum number of moves to satisfy the above requirement.

Input

The first line of input contains two integers $n$ and $m$ ($1\le n\le 2\cdot {10}^5,1\le m\le n$). It is guaranteed that $m$ is a divisor of $n$.

The second line of input contains $n$ integers $a_1,a_2,\dots ,a_n$ ($0\le a_i\le {10}^9$), the elements of the array.

Output

In the first line, print a single integer — the minimum number of moves required to satisfy the following condition: for each remainder from $0$to $m-1$, the number of elements of the array having this remainder equals $\frac{n}{m}$.

In the second line, print any array satisfying the condition and can be obtained from the given array with the minimum number of moves. The values of the elements of the resulting array must not exceed ${10}^{18}$.

Examples

input

Copy

6 3
3 2 0 6 10 12

output

Copy

3
3 2 0 7 10 14 

input

Copy

4 2
0 1 2 3

output

Copy

0
0 1 2 3 

题意：本题就是给你n，m,保证n能被m整除，给你n个数，对这些数操作+=1，使得这些数%m后，得到的数是从0~m-1，且没个数出现n/m次。

题解：贪心，对于数量少的先不处理，对于多于n/m的数使其变为离他最近的数量不到n/m的数，记录需要操作的次数，跑一边就可以得到结果；

AC代码为：

#include
using namespace std;
const int maxn=2e5+10;
const int INF=0x3f3f3f3f;
typedef long long LL;


vector v[maxn];
int a[maxn],n,m;
LL ans;


int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>m;ans=0;
    
    for(int i=0;i    {
        cin>>a[i];
        v[a[i]%m].push_back(i);
    }
    
    int temp=0,flag=n/m;
    for(int i=0;i    {
        while(v[i].size() > flag)
        {
            temp=max(temp,i);
            while(v[temp%m].size()>=flag) temp++;
            int num=min(flag-v[temp%m].size(),v[i].size()-flag);
            int c_num=temp-i;
            while(num--)
            {
                ans+=c_num;
                a[v[i].back()]+=c_num;
                v[temp%m].push_back(v[i].back());
                v[i].pop_back();
            }
        }
    }
    
    cout<    for(int i=0;i    
    return 0;
}

转载于:https://www.cnblogs.com/songorz/p/9386536.html