题干：
You are given an array consisting of nn integers a1,a2,…,ana1,a2,…,an, and a positive integer mm. It is guaranteed that mm is a divisor of nn.
In a single move, you can choose any position ii between 11 and nn and increase aiai by 11.
Let's calculate crcr (0≤r≤m−1)0≤r≤m−1) — the number of elements having remainder rr when divided by mm. In other words, for each remainder, let's find the number of corresponding elements in aa with that remainder.
Your task is to change the array in such a way that c0=c1=⋯=cm−1=nmc0=c1=⋯=cm−1=nm.
Find the minimum number of moves to satisfy the above requirement.
Input
The first line of input contains two integers nn and mm (1≤n≤2⋅105,1≤m≤n1≤n≤2⋅105,1≤m≤n). It is guaranteed that mm is a divisor of nn.
The second line of input contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤1090≤ai≤109), 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 00 to m−1m−1, the number of elements of the array having this remainder equals nmnm.
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 10181018.
Examples
Input
6 3 3 2 0 6 10 12
Output
3 3 2 0 7 10 14
Input
4 2 0 1 2 3
Output
0 0 1 2 3
题目大意：
给定n和m（保证m是n的因子），给出n个数的序列，定义一次操作：任选序列中的一个位置，使这个位置上的数加一。定义c[x]代表a[i]%m==x的数的个数
问你最少需要多少次操作，可以使得c[0]=c[1]=c[2]=...=c[m-1]=n/m。
(n<=2e5，a[i]<=1e18)
解题报告：
刚开始想了一个比较麻烦的做法：首先对所有数都%m，因为除了最后输出的需要，其他时候a[i]再大都没啥作用。然后用两个map分别维护哪些数的个数不够，哪些数的个数多了，然后贪心的搞。但是这样有个问题，不方便记录它对应的是哪些数。也就意味着不方便输出原序列。如果题目只要求输出最少次数的话，就可以这样做了。
题解是这样做的：思想还是把插入的问题转化成删除的问题。先把所有余数都加到一个set中，然后每读入一个数a[i]，都从set中找距离他最近的数，并且让他变到这个数。如果x这个数已经够了n/m次了，那就从set中删除。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
set<ll> ss;
ll a[MAX],b[MAX],cnt[MAX],n,m,ci,ans;
int main()
{cin>>n>>m;ci=n/m;for(int i = 0; i<m; i++) ss.insert(i); for(int i = 1; i<=n; i++) scanf("%lld",a+i),b[i] = a[i]%m;for(int tmp,i = 1; i<=n; i++) {if(b[i] > *ss.rbegin()) tmp = *ss.begin();else tmp = *ss.lower_bound(b[i]);cnt[tmp]++;if(cnt[tmp] == ci) ss.erase(tmp);ans += (m+tmp-b[i])%m;a[i] += (m+tmp-b[i])%m;}printf("%lld\n",ans);for(int i = 1; i<=n; i++) printf("%lld ",a[i]);return 0 ;
}
