Bzoj 2568-bit collection

Title Link: http://www.lydsy.com:808/JudgeOnline/problem.php?id=2568

Test instructions: Maintains a collection s that supports the following operations:

(1) INS m: inserting the element m into the set S;
(2) DEL M: Removes all elements equal to M in the set S;
(3) Add M: Adds the value m to all elements in the set S;
(4) Qbit K: The number of elements in the query collection satisfies the second binary's K-bit is 1.

Idea: I'm looking at this.

https://github.com/strongoier/OJ/tree/master/BZOJ/2568%20%E6%AF%94%E7%89%B9%E9%9B%86%E5%90%88

The general idea of the original is as follows:

(1) The ADD operation of the one and separate out, set to sum, the set S of each element x the actual value is x+sum;

(2) Set F[k][t] indicates that the K bit is 1, and this number is less than the number of digits equal to D. Each query, set l=2^x,r=2^ (x+1)-1, the answer is F[K][R]-F[K][L-1];

(3) This f value is a segment change due to the addition of a number, so a tree array is used to maintain the F-array;

(4) for the number of inserted numbers and how many each number, with a map record, so that the deletion will know how much to subtract in the tree array.

I had a problem that I did not understand, because the actual number to insert when inserting x is X-sum, then x-sum is negative when this bit is not the same as a positive digit. The binary representation of a negative number is the binary representation of the corresponding positive number and the negation plus 1.

Like what

-1=11111111 11111111 11111111 11111111

-2=11111111 11111111 11111111 11111110

-3=11111111 11111111 11111111 11111101

-4=11111111 11111111 11111111 11111100

The insertion of the great God above is directly inserted

There's a 1 in the back because the subscript in the tree array starts at 1.

And then the sum is like this.

This is divided into two parts, the first part: Calculate the [l,r] interval, set k=2, then the binary representation l=100,r=111. Set sum=1011, then the actual interval to be calculated is [001,100], as long as a number of the last three bits in this interval, that is [001,100], then it plus sum after the three will fall to the [l,r] interval. In fact, this is not carry.

We set the sum=1110 again, the other constant, then the actual summation interval above becomes [000,001]. We found that, in addition to this interval, [110,111] This range is also possible. This is actually a carry, after the sum interval from [100,111] to [1100,1111], so minus the sum of the next three bits 110 the actual interval is [110,1001], we found that 1001,1000 will not have this value, so the actual is [110,111]. This is the second part of the summation above.

Then a negative number plus sum can also reach this interval, sum=1110,[-10,-7], the binary of these negative numbers is

-10=11111111 11111111 11111111 11110110

-9 = 11111111 11111111 11111111 11110111

-8 = 11111111 11111111 11111111 11111000

-7 = 11111111 11111111 11111111 11111001

We found that the latter three were in the two intervals of the calculation. So negative numbers don't need extra consideration.

int S[20][n];map<int,int> MP; int n;  void Add (int k,int X,int t) {while (x<n) s[k][x]+=t,x+=x&-x;}    int get (int k,int x) {int ans=0;    while (x) ans+=s[k][x],x-=x&-x; return ans;}     int main () {n=myint ();    int sum=0;        while (n--) {char op[10];        int x;        scanf ("%s%d", op,&x);        if (' A ' ==op[0]) sum+=x;            else if (' I ' ==op[0]) {x-=sum;            mp[x]++;        for (int i=0;i<16;i++) Add (I, (x& ((1<< (i+1))-1)) +1,1);            } else if (' D ' ==op[0]) {x-=sum;            int t=mp[x];            mp[x]=0;        for (int i=0;i<16;i++) Add (I, (x& ((1<< (i+1))-1) +1,-t);            } else if (' Q ' ==op[0]) {int ans=0;            int l=1<<x,r= (1<< (x+1))-1;            Ans+=get (X,min (1<<16,max (0,r-(sum& (1<<)-1))));            Ans-=get (X,min (1<<16,max (0,l-(sum& (1<<)-1)))); l|=1<< (X+1);            r|=1<< (x+1);            Ans+=get (X,min (1<<16,max (0,r-(sum& (1<<)-1))));            Ans-=get (X,min (1<<16,max (0,l-(sum& (1<<)-1))));        printf ("%d\n", ans); }    }}

　　

Bzoj 2568-bit collection