数列 暴力 贪心+剪枝

Training 2 - F题

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][0,0,1,0,2]

MEX equals to 33

because numbers 0,10,1

and 22

are presented in the array and 33

is the minimum non-negative integer not presented in the array; for the array [1,2,3,4][1,2,3,4]

MEX equals to 00

because 00

is the minimum non-negative integer not presented in the array; for the array [0,1,4,3][0,1,4,3]

MEX equals to 22

because 22

is the minimum non-negative integer not presented in the array. 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

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]a=[0]

you don’t need to perform any operations, maximum possible MEX is 11

. After the second query, the array is a=[0,1]a=[0,1]

you don’t need to perform any operations, maximum possible MEX is 22

. After the third query, the array is a=[0,1,2]a=[0,1,2]

you don’t need to perform any operations, maximum possible MEX is 33

. After the fourth query, the array is a=[0,1,2,2]a=[0,1,2,2]

you don’t need to perform any operations, maximum possible MEX is 33

(you can’t make it greater with operations). After the fifth query, the array is a=[0,1,2,2,0]a=[0,1,2,2,0]

you can perform a[4]:=a[4]+3=3a[4]:=a[4]+3=3

. The array changes to be a=[0,1,2,2,3]a=[0,1,2,2,3]

. Now MEX is maximum possible and equals to 44

. After the sixth query, the array is a=[0,1,2,2,0,0]a=[0,1,2,2,0,0]

you can perform a[4]:=a[4]+3=0+3=3a[4]:=a[4]+3=0+3=3

. The array changes to be a=[0,1,2,2,3,0]a=[0,1,2,2,3,0]

. Now MEX is maximum possible and equals to 44

. After the seventh query, the array is a=[0,1,2,2,0,0,10]a=[0,1,2,2,0,0,10]

. You can perform the following operations: a[3]:=a[3]+3=2+3=5a[3]:=a[3]+3=2+3=5

, a[4]:=a[4]+3=0+3=3a[4]:=a[4]+3=0+3=3

, a[5]:=a[5]+3=0+3=3a[5]:=a[5]+3=0+3=3

, a[5]:=a[5]+3=3+3=6a[5]:=a[5]+3=3+3=6

, a[6]:=a[6]−3=10−3=7a[6]:=a[6]−3=10−3=7

, a[6]:=a[6]−3=7−3=4a[6]:=a[6]−3=7−3=4

. The resulting array will be a=[0,1,2,5,3,6,4]a=[0,1,2,5,3,6,4]

. Now MEX is maximum possible and equals to 77

.

#pragma warning (disable:4996)
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define inf 0X3f3f3f3f
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 4e5 + 5;

int times[maxn];//每个数出现的次数
bool v[maxn];//记录当前数字是否已经存在

int main()
{
	int q, x;
	scanf("%d%d", &q, &x);
	if (q == 0)
		printf("0\n");
	int ans = 0;
	for (int i = 1; i <= q; i++)
	{
		int y;
		scanf("%d", &y);

		ull tmp = y % x + (ull)times[y % x] * x;
		if (tmp <= q)
			v[tmp] = 1;
		times[y % x]++;

		while (v[ans] == 1)
			ans++;

		printf("%d\n", ans);

	}
	return 0;
}

思路：
对于每一个数字y可以拆分成y%x+n*x，所以统计y%x出现的次数，同时记录每一个数字是否已经被填满。
时间卡的很近，数据要用ull，还需要判断每一次的数值是不是大于q。

%大佬的代码