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Codeforces Round #371 (Div. 2) C. Sonya and Queries —— 二进制压缩

發布時間:2023/12/20 编程问答 35 豆豆
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題目鏈接:http://codeforces.com/contest/714/problem/C


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?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.



  • 題解:?

    由于查詢模式中只有01串(代表奇偶),所以輸入的數字只需用01記錄每一位奇偶性。故可以用二進制對數字進行壓縮歸類。由于1<<18=26214, 范圍不是很大,所以可以用數組記錄每個二進制數出現的個數,之后就可以直接修改,查詢了。



    代碼如下:

    #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<cmath> #include<queue> #include<vector> #include<map> #include<string> #include<set> #define LL long long #define MAX(a,b) (a>b?a:b) #define MIN(a,b) (a<b?a:b) #define INF 0x7fffffff #define LNF ((1LL<<62)-1) #define mod 1000000007 #define maxn 300000using namespace std;LL p[maxn]; char ch, s[50];int main() {int n,bin;scanf("%d",&n);for(int i = 0; i<n; i++){getchar();scanf("%c%s",&ch,s);bin = 0;for(int j = 0,len = strlen(s); j<len; j++){bin <<= 1;if((s[j]-'0')&1)bin++;}if(ch=='+')p[bin]++;else if(ch=='-')p[bin]--;elseprintf("%lld\n",p[bin]);} }

    轉載于:https://www.cnblogs.com/DOLFAMINGO/p/7538740.html

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