Equalize the Remainders（set二分+思维）

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
题意：构造出这样的一组序列，使得这个数组中的每个数字取模m之后的数字即0~m-1都出现过n/m次。问最少需要操作多少次。
思路：对于余数出现次数少于n/m次的数，我们放到set中去。
对于出现次数多于n/m的数，我们二分去找set中第一个大于这个数的数。如果操作之后这个数的出现次数等于n/m次了，就将之从集合中删除。
代码如下：

#include
#define ll long long
using namespace std;

const int maxx=2e5+100;
ll a[maxx];
int b[maxx];
int n,m;

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) 
	{
		scanf("%I64d",&a[i]);
		b[a[i]%m]++;
	}
	set<ll> s;
	for(int i=0;i<m;i++)
	{
		if(b[i]<n/m) s.insert(i);
	}
	ll ans=0;
	for(int i=1;i<=n;i++)
	{
		if(b[a[i]%m]<=n/m) continue;
		b[a[i]%m]--;
		set<ll> ::iterator it=s.lower_bound(a[i]%m);
		if(it!=s.end())//找到了
		{
			ans+=*it-a[i]%m;
			a[i]+=*it-a[i]%m;
			b[*it]++;
			if(b[*it]==n/m) s.erase(it);
		}
		else//没有找到的话，就要变成集合中的第一个元素。
		{
			ans+=(*s.begin()+m-a[i]%m)%m;
			a[i]+=(*s.begin()+m-a[i]%m)%m;
			b[*s.begin()]++;
			if(b[*s.begin()]==n/m) s.erase(s.begin());
		}	
	}
	cout<<ans<<endl;
		for(int i=1;i<=n;i++) cout<<a[i]<<" ";
		cout<<endl;
}

努力加油a啊，(^o)/~