Codeforces Round #615 (Div. 3)D. MEX maximizing（取模找最小值）

D. MEX maximizing

time limit per test3 seconds
memory limit per test256 megabytes

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

for the array [0,0,1,0,2] MEX equals to 3 because numbers 0,1 and 2
are presented in the array and 3 is the minimum non-negative integer
not presented in the array; for the array [1,2,3,4] MEX equals to 0
because 0 is the minimum non-negative integer not presented in the
array; for the array [0,1,4,3] MEX equals to 2 because 2 is the
minimum non-negative integer not presented in the array. You are given
an empty array a=[] (in other words, a zero-length array). You are
also given a positive integer x.

You are also given q queries. The j-th query consists of one integer
yj and means that you have to append one element yj to the array. The
array length increases by 1 after a query.

In one move, you can choose any index i and set ai:=ai+x or ai:=ai−x
(i.e. increase or decrease any element of the array by x). The only
restriction is that ai 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 q queries (i.e. the j-th
answer corresponds to the array of length j).

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

Input

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

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

Output

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

Examples

input

7 3
0
1
2
2
0
0
10

output

1
2
3
3
4
4
7

input

4 3
1
2
1
2

output

0
0
0
0

Note

In the first example:

After the first query, the array is a=[0]: you don’t need to perform
any operations, maximum possible MEX is 1. After the second query, the
array is a=[0,1]: you don’t need to perform any operations, maximum
possible MEX is 2. After the third query, the array is a=[0,1,2]: you
don’t need to perform any operations, maximum possible MEX is 3. After
the fourth query, the array is a=[0,1,2,2]: you don’t need to perform
any operations, maximum possible MEX is 3 (you can’t make it greater
with operations). After the fifth query, the array is a=[0,1,2,2,0]:
you can perform a[4]:=a[4]+3=3. The array changes to be a=[0,1,2,2,3].
Now MEX is maximum possible and equals to 4. After the sixth query,
the array is a=[0,1,2,2,0,0]: you can perform a[4]:=a[4]+3=0+3=3. The
array changes to be a=[0,1,2,2,3,0]. Now MEX is maximum possible and
equals to 4. After the seventh query, the array is a=[0,1,2,2,0,0,10].
You can perform the following operations: a[3]:=a[3]+3=2+3=5,
a[4]:=a[4]+3=0+3=3, a[5]:=a[5]+3=0+3=3, a[5]:=a[5]+3=3+3=6,
a[6]:=a[6]−3=10−3=7, a[6]:=a[6]−3=7−3=4. The resulting array will be
a=[0,1,2,5,3,6,4]. Now MEX is maximum possible and equals to 7.

AC代码：

#pragma GCC optimize(3,"Ofast","inline")
#include<iostream>

using namespace std;
const int maxn = 400010;
int a[maxn];

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int q, x, y;
	int i = 0;
	cin >> q >> x;
	while (q--)
	{
		cin >> y;
		a[y % x]++;
		while (a[i % x])
		{
			a[i % x]--;
			i++;
		}
		cout << i << endl;
	}
	return 0;
}