題干：

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

題目大意：

給定n和m（保證m是n的因子），給出n個數的序列，定義一次操作：任選序列中的一個位置，使這個位置上的數加一。定義c[x]代表a[i]%m==x的數的個數

問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)

解題報告：

剛開始想了一個比較麻煩的做法：首先對所有數都%m，因為除了最后輸出的需要，其他時候a[i]再大都沒啥作用。然后用兩個map分別維護哪些數的個數不夠，哪些數的個數多了，然后貪心的搞。但是這樣有個問題，不方便記錄它對應的是哪些數。也就意味著不方便輸出原序列。如果題目只要求輸出最少次數的話，就可以這樣做了。

題解是這樣做的：思想還是把插入的問題轉化成刪除的問題。先把所有余數都加到一個set中，然后每讀入一個數a[i]，都從set中找距離他最近的數，并且讓他變到這個數。如果x這個數已經夠了n/m次了，那就從set中刪除。

AC代碼：

#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 ; }

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