原题：
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
inputCopy
7 3
0
1
2
2
0
0
10
outputCopy
1
2
3
3
4
4
7
inputCopy
4 3
1
2
1
2
outputCopy
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.
题意：
q次询问，每次放进一个数（开始为空数组），可以对数组内的某个数或者某几个数做任意操作数的加x或者减x （x为input第二个数），处理完后询问，在该数组情况下的最小整数，就是数组里的数都被排除的情况下，从0开始找，第一个不在数组里的整数就是了，然后要求操作加x或者减x的任意操作后使得那个询问结果最大化。
思路：水题啊！！！明显是要让他尽可能覆盖从0开始的整数，把小的整数盖掉才会得到大的结果，如果小的那个比如说1，已经被覆盖，然后给你一个数它是1+nx，毫无疑问这个数%x得1，然后从1开始找，1已经有了，就1+1x覆盖，以此类推就行了！本题问题主要是方法乱容易超时，数据大容易runtime error。
题目链接
#include<iostream>
#include<vector>
#include<set>
#include<string>
#include<map>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string>
#include<iomanip>
#include<queue>
#include<vector>
#include<set>
#include<cstring>
using namespace std;
int f[400005];
bool a[400005];
int main (){
int q,x,ans=0;
scanf("%d%d",&q,&x);
for(int i=1;i<=q;i++){
int y;
scanf ("%d",&y);
y%=x;
unsigned long long dd=y+1*x*f[y];//无unsigned会喜提runtime error
if(dd<=q)
a[dd]=true;
f[y]++;//f[y]是%出来的余数的位置，记录这个体系的出现了几次，然后可以直接乘，就省时间
while(a[ans])
ans++;
printf("%d\n",ans);
}
return 0;
}