Codeforces Round #371 (Div. 2) C. Sonya and Queries[Map

Codeforces Round #371 (Div. 2) C. Sonya and Queries[Map｜二进制]

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发布时间：2019-06-29

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Today Sonya learned about long integers and invited all her friends to share the fun. Sonya has an initially empty multiset with integers. Friends give her t queries, each of one of the following type:

+ ai — add non-negative integer ai to the multiset. Note, that she has a multiset, thus there may be many occurrences of the same integer.

- ai — delete a single occurrence of non-negative integer ai from the multiset. It's guaranteed, that there is at least one ai in the multiset.

? s — count the number of integers in the multiset (with repetitions) that match some pattern s consisting of 0 and 1. In the pattern, 0stands for the even digits, while 1 stands for the odd. Integer x matches the pattern s, if the parity of the i-th from the right digit in decimal notation matches the i-th from the right digit of the pattern. If the pattern is shorter than this integer, it's supplemented with 0-s from the left. Similarly, if the integer is shorter than the pattern its decimal notation is supplemented with the 0-s from the left.

For example, if the pattern is s = 010, than integers 92, 2212, 50 and 414 match the pattern, while integers 3, 110, 25 and 1030 do not.

Input

The first line of the input contains an integer t (1 ≤ t ≤ 100 000) — the number of operation Sonya has to perform.

Next t lines provide the descriptions of the queries in order they appear in the input file. The i-th row starts with a character ci — the type of the corresponding operation. If ci is equal to '+' or '-' then it's followed by a space and an integer ai (0 ≤ ai < 1018) given without leading zeroes (unless it's 0). If ci equals '?' then it's followed by a space and a sequence of zeroes and onse, giving the pattern of length no more than 18.

It's guaranteed that there will be at least one query of type '?'.

It's guaranteed that any time some integer is removed from the multiset, there will be at least one occurrence of this integer in it.

Output

For each query of the third type print the number of integers matching the given pattern. Each integer is counted as many times, as it appears in the multiset at this moment of time.

Examples

input

12 + 1 + 241 ? 1 + 361 - 241 ? 0101 + 101 ? 101 - 101 ? 101 + 4000 ? 0

output

2 1 2 1 1

input

4 + 200 + 200 - 200 ? 0

output

Note

Consider the integers matching the patterns from the queries of the third type. Queries are numbered in the order they appear in the input.

1 and 241.

361.

101 and 361.

361.

4000.

题意：插入删除数字查询模式

一开始想了一个好像很厉害的做法，位向量树，实在调试不出来了

然后突然发现，数字是按十进制不是二进制

并且，数字和模式高位的0补位其实没什么影响，就是没有值，就是没有

所有把数字和模式都变成十进制或二进制01串然后用map或一个数组(2^18=262144)加减就行了

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          using namespace std;typedef long long ll;const int N=500000;inline ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){
    if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}ll t,a;char c[3],s[20];//map
           
             mp;int mp[N];int main(int argc, const char * argv[]) { t=read(); while(t--){ scanf("%s",c);a=read(); ll b=0,k=1; while(a){ b+=a%2*k; a/=10; k*=2; } if(c[0]=='+') mp[b]++; if(c[0]=='-') mp[b]--; if(c[0]=='?') printf("%d\n",mp[b]); } return 0;}

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