Codeforces #317(div.2) C(二进制 状压)
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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 outputToday 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 1030 do not.
InputThe 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.
OutputFor 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
Input12
+ 1
+ 241
? 1
+ 361
- 241
? 0101
+ 101
? 101
- 101
? 101
+ 4000
? 0Output
2
1
2
1
1Input
4
+ 200
+ 200
- 200
? 0Output
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.
分析
刚开始,还以为要用类似于哈夫曼树的写法,将每个数的特征表示出来,查询的时候,dfs一遍节点就行了。
后来,发现想多了,只要 利用二进制的特点, 压缩每一个数就行了。
代码
/*===============================================================* Copyright (C) 2016 All rights reserved.* * Filename: C.cpp* Author: gsh* Created_time:2016/9/16 星期五 15:03:27* Problem_tags:* Problem_keyword:================================================================*/#include <bits/stdc++.h>typedef long long ll;typedef unsigned long long ull;using namespace std;const int maxn = 1000010; int ans[maxn]; int main() { int n; long long a; char ch; scanf("%d",&n); memset(ans,0,sizeof(ans)); while(n--) { scanf(" %c %I64d",&ch, &a); int sum = 0, k = 1; while(a) { sum += (a%2)*k; a /= 10; k *= 2; } if(ch == '+') ans[sum]++; if(ch == '-') ans[sum]--; if(ch == '?') cout << ans[sum] << endl; } return 0; }
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