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

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C. Sonya and Queries time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

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,0 stands 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 with0-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 1 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.

题目大意：给你一个mutiset，每次询问一个01串s，如果s的某一位为0，那么这一位所匹配的是偶数，若为1则匹配奇数，若匹配时产生数位不够则在前导加0.问每次询问有多少个数与s匹配。

解题思路：因为偶数是与0匹配的，在二进制中前导为0的话可以无视，那么我们可以以奇数为边界，因为奇数与1匹配，。

所以，我们把插进来的数的偶数前导去掉，然后用map搞搞就能够统计结果了。

/* *********************************************** ┆ ┏┓　　　┏┓ ┆ ┆┏┛┻━━━┛┻┓ ┆ ┆┃　　　　　　　┃ ┆ ┆┃　　　━　　　┃ ┆ ┆┃　┳┛　┗┳　┃ ┆ ┆┃　　　　　　　┃ ┆ ┆┃　　　┻　　　┃ ┆ ┆┗━┓　马　┏━┛ ┆ ┆　　┃　勒　┃　　┆　　　　　　 ┆　　┃　戈　┗━━━┓ ┆ ┆　　┃　壁　　　　　┣┓┆ ┆　　┃　的草泥马　　┏┛┆ ┆　　┗┓┓┏━┳┓┏┛ ┆ ┆　　　┃┫┫　┃┫┫ ┆ ┆　　　┗┻┛　┗┻┛ ┆ ************************************************ */ #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <stack> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <bitset> using namespace std; #define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++) #define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--) #define pb push_back #define mp make_pair const int inf_int = 2e9; const long long inf_ll = 2e18; #define inf_add 0x3f3f3f3f #define mod 1000000007 #define LL long long #define ULL unsigned long long #define MS0(X) memset((X), 0, sizeof((X))) #define SelfType int SelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);} SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;} #define Sd(X) int (X); scanf("%d", &X) #define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y) #define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z) #define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin()) #define all(a) a.begin(), a.end() typedef pair<int, int> pii; typedef pair<long long, long long> pll; typedef vector<int> vi; typedef vector<long long> vll; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;} //#pragma comment(linker, "/STACK:102400000,102400000") map<LL,int>p; LL res[20]; LL change(LL x) { for(int i=0;i<18;i++) { res[i] = x % 10; res[i] &= 1; x /= 10; } LL ans = 0; for(int i=18;i>=0;i--) { ans *= 10; ans += res[i]; } return ans; } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); ios::sync_with_stdio(0); cin.tie(0); int t; t = read(); while(t--) { LL x; char s[10]; scanf("%s%I64d",s,&x); if(s[0]=='+') { x = change(x); p[x]++; } else if(s[0]=='-') { x = change(x); p[x]--; } else { if(p.count(x))printf("%d\n",p[x]); else printf("0\n"); } } return 0; }