Codeforces #371（Div.2）C.Sonya and Queries【map应用】【思维】

传送门 ：C. Sonya and Queries

描述：

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 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 1030do 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.

题意：

一个multiset，有t个操作，其中+ a表示multiset中加入一个数a，-a表示从multiset取出a，?表示每次询问一个01串s，如果s的一位是0，那么所匹配的数的该位应该是偶数，反之如果是1所匹配的改位应该是奇数。如果匹配时产生数位不够的问题的话添加前导0。每次询问有多少个数和s串能够匹配。

这题意要看半天_(:зゝ∠)_

思路：

不是题目中有multiset，就一定要用multiset，这题其实就是题目难读，仔细思考一下用map操作一下就好了

+a时将 a变成一个对应询问的01串即可。举例说明：

361==101

241==001==1

然后对应保存这个得到的01串（要用long long ）为tmp，然后map[tmp]++；

同理-a 对应着map[tmp]--；

那么在询问的时候，直接输出当前01串映射的值即可。

代码：

#include <bits/stdc++.h> #define ll __int64 using namespace std; map<ll, int>mp; ll change(ll a){ ll res=0,t=1; while(a){ res+=(a%2)*t; t*=10; a/=10; } return res; } int main(){ int t; while(~scanf("%d",&t)){ mp.clear(); char s[4];ll a; while(t--){ scanf("%s%I64d",s,&a); ll tmp=change(a); if(s[0]=='+')mp[tmp]++; else if(s[0]=='-')mp[tmp]--; else printf("%d\n",mp[tmp]); } } return 0; }

另附某大牛的做法：

       基本思想差不多，就是转换成01串时，再把二进制的01串转换成10进制的数字

#include <bits/stdc++.h> using namespace std; int cnt[1<<18]; int main() { int t; scanf("%d",&t); while(t--) { char ty[5],buf[25]; scanf("%s%s",ty,buf); int tmp=0; for(int i=0;buf[i];i++) tmp=tmp*2+(buf[i]-'0')%2; if(*ty=='?')printf("%d\n",cnt[tmp]); else cnt[tmp]+=(*ty=='+' ? 1 : -1); } return 0; } PS:大牛和若菜的区别QAQ,不仅要能做对，还要做的更好，代码更优更少。。。。