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.
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.
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.
12
+ 1
+ 241
? 1
+ 361
- 241
? 0101
+ 101
? 101
- 101
? 101
+ 4000
? 0
2
1
2
1
1
4
+ 200
+ 200
- 200
? 0
1
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.
題意：插入刪除數字 查詢模式
一開始想了一個好像很厲害的做法，位向量樹，實在除錯不出來了
然後突然發現，數字是按十進位制不是二進位制
並且，數字和模式高位的0補位其實沒什麼影響，就是沒有值，就是沒有
所有把數字和模式都變成十進位制或二進位制01串然後用map或一個陣列(2^18=262144)加減就行了
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <map> 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<ll,int> 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; }