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D. MEX maximizing

作者:互联网

Recall that MEX of an array is a minimum non-negative integer that does not belong to the array. Examples:

You are given an empty array a=[]a=[] (in other words, a zero-length array). You are also given a positive integer xx.

You are also given qq queries. The jj-th query consists of one integer yjyj and means that you have to append one element yjyj to the array. The array length increases by 11 after a query.

In one move, you can choose any index ii and set ai:=ai+xai:=ai+x or ai:=ai−xai:=ai−x (i.e. increase or decrease any element of the array by xx). The only restriction is that aiai cannot become negative. Since initially the array is empty, you can perform moves only after the first query.

You have to maximize the MEX (minimum excluded) of the array if you can perform any number of such operations (you can even perform the operation multiple times with one element).

You have to find the answer after each of qq queries (i.e. the jj-th answer corresponds to the array of length jj).

Operations are discarded before each query. I.e. the array aa after the jj-th query equals to [y1,y2,…,yj][y1,y2,…,yj].

Input

The first line of the input contains two integers q,xq,x (1≤q,x≤4⋅1051≤q,x≤4⋅105) — the number of queries and the value of xx.

The next qq lines describe queries. The jj-th query consists of one integer yjyj (0≤yj≤1090≤yj≤109) and means that you have to append one element yjyj to the array.

Output

Print the answer to the initial problem after each query — for the query jj print the maximum value of MEX after first jj queries. Note that queries are dependent (the array changes after each query) but operations are independent between queries.

Examples input Copy
7 3
0
1
2
2
0
0
10
output Copy
1
2
3
3
4
4
7
input Copy
4 3
1
2
1
2
output Copy
0
0
0
0
Note

In the first example:

 

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>
//#include <xfunctional>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define pb              push_back
#define mk              make_pair
using namespace std;
int dir1[6][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 },{ 1,1 },{ -1,1 } };
int dir2[6][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 },{ 1,-1 },{ -1,-1 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979;
const int mod = 100007;
const int N = 1005;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}

int main()
{
    int q, x,ans=0;
    cin >> q >> x;
    vector<int> a(q+1,0);
    while (q--)
    {
        int t;
        cin >> t;
        a[t%x]++;
        while (a[ans%x])
        {
            a[ans%x]--;
            ans++;
        }
        cout << ans << endl;
    }
    return 0;
}

 

标签:operations,maximizing,perform,query,array,include,MEX
来源: https://www.cnblogs.com/dealer/p/12785215.html