[原创]codeforces 714C Sonya and Queries [思维]【STL】

2016-09-14 14:25:00 Tabris_ 阅读数：700

博客爬取于2020-06-14 22:43:19
以下为正文

版权声明：本文为Tabris原创文章，未经博主允许不得私自转载。
https://blog.csdn.net/qq_33184171/article/details/52537683

题目链接:http://codeforces.com/contest/714/problem/C

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C. Sonya and Queries
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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.
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题目大意：（懂不懂题目的我眼泪掉下来~~~）
就是有一个multiset 这么一个容器
有三种操作

a 把a加入multiset

a 把a删除multiset
？s 寻找s是一个<=18位的省去了前导0的二进制数, 然后求当前multiset中每位%2后等于s的数的个数

解题思路：
就是维护一个map
映射的事这个状态下的个数
然后就能在O（18n）的复杂度中求解出来了

附本题代码
----------------------------.

# include <bits/stdc++.h>
using namespace std;

# define LL long long int
# define INF 0x1f1f1f1f
# define pb push_back

# define lalal puts("*******");
/*************************************/

const int MOD = 1e9+7;
const int M = 1e5+10;

map<LL,int>b;
int main()
{
    ios::sync_with_stdio(false);
    int t;
    while(cin>>t)
    {
        b.clear();
        LL tem ,x,a,bit;
        char c;
        for(int i=0; i<t; i++)
        {
            cin>>c>>a;
            tem = a,x = 0,bit=0;
            while(tem)
            {
                if(tem%10%2)    x  |=  (1<<bit);
                tem/=10;
                bit++;
            }
            if(c=='+') b[x]++;
            if(c=='-') b[x]--;

            if(c=='?') cout<<b[x]<<endl;
        }
    }
    return 0;
}