CodeForces - 999D - Equalize the Remainders

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

[分析]
遍历每个数字，找到最接近的&&没有满的余数。
类似贪心
所以先将没满的余数都放入set
如果这个数字满了，就把他从set中去除。
因为set是已经排序了得！！！
这一点很重要。
我们就可以在set中（在set中说明还没满）用二分查找出第一个大于等于a[i]%m的合法余数
如何计算得到移动次数和a[i]就很简单了

总结，需要关注set的排序性，和二分

#include
#include
using namespace std;
using namespace std;
#define MAXN 200005

long long a[MAXN], cnt[MAXN], ans;
set<int> s;
int main()
{
    int n, m;

    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; i++)
    {
        s.insert(i);
    }
    for (int i = 0; i < n; i++)
    {
        scanf("%lld", &a[i]);
        int now = a[i] % m;
        int x;
        if (now > *s.rbegin()) x = *s.begin();
        else x = *s.lower_bound(now);
        if (++cnt[x] == n / m) s.erase(x);
        ans += (x - now + m) % m;
        a[i] += (x - now + m) % m;
    }
    printf("%lld\n", ans);
    for (int i = 0; i < n; i++) printf("%lld ", a[i]);
}