# codeforces 999D 贪心+二分查找

xiaoxiao2021-02-28  37

D. 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 (0rm1)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==cm1=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 (1n2105,1mn1≤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 (0ai1090≤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 m1m−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 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

#include<stdio.h> #include<string.h> #include<cmath> #include<stdlib.h> #include<time.h> #include<algorithm> #include<iostream> #include<vector> #include<queue> #include<set> #define ll long long #define qq printf("QAQ\n"); using namespace std; const int maxn=1e5+5; const int inf=2e9+1e8+1234; const ll linf=8e18+9e17; const int mod=1e9+7; const double e=exp(1.0); const double pi=acos(-1); int a[maxn<<1],b[maxn<<1]; bool use[maxn<<1]={0}; int main() { int n,m; while(scanf("%d%d",&n,&m)!=EOF) { memset(a,0,sizeof a); memset(b,0,sizeof b); int t=n/m; ll ans=0; //vector<int>v; //cout<<v.size()<<endl; set<int>s; set<int>::iterator iter; for(int i=0;i<m;i++) /*v.push_back(i),*/s.insert(i); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); int temp=a[i]%m; if(b[temp]<t){ b[temp]++; //if(b[temp]==t)v.erase(lower_bound(v.begin(),v.end(),temp)); if(b[temp]==t)s.erase(temp); } else { //int f=lower_bound(v.begin(),v.end(),temp)-v.begin(); iter=s.lower_bound(temp); //if(f+v.begin()==v.end()){ if(iter==s.end()){ //a[i]+=m-temp+v[0]; a[i]+=m-temp+*s.begin(); //ans+=(ll)(m-temp+v[0]); ans+=(ll)(m-temp+*s.begin()); b[a[i]%m]++; //if(b[a[i]%m]==t)v.erase(lower_bound(v.begin(),v.end(),a[i]%m)); if(b[a[i]%m]==t)s.erase(a[i]%m); } else { //ans+=(ll)(v[f]-a[i]%m); ans+=(ll)(*iter-a[i]%m); //a[i]+=v[f]-temp; a[i]+=*iter-temp; //b[v[f]]++; b[*iter]++; //if(b[a[i]%m]==t)v.erase(lower_bound(v.begin(),v.end(),a[i]%m)); if(b[a[i]%m]==t)s.erase(a[i]%m); } } } printf("%lld\n",ans); for(int i=1;i<=n;i++) if(i==n)printf("%d\n",a[i]); else printf("%d ",a[i]); } return 0; } /* 6 2 1 1 1 1 1 1 6 3 3 2 0 6 10 12 */