Cattle off more school sixth -H-Pair

Links: https://ac.nowcoder.com/acm/contest/887/H
Source: Cattle-off network

Title Description

Given three integers A, B, C. Count the number of pairs < x ,y> (with 1≤x≤Aand1≤y≤B)
such that at least one of the following is true:
- (x and y) > C
- (x xor y) < C
 
("and", "xor" are bit operators)

Enter a description:

The first line of the input gives the number of test cases,T(T≤100). T test cases follow.

For each test case, the only line contains three integers A, B and C.
1≤A,B,C≤10^9

Output Description:

For each test case, the only line contains an integer that is the number of pairs satisfying the condition given in the problem statement.

Example 1

Entry

3
3 4 2
4 5 2
7 8 5

Export

5
7
31

Digital dp request x & y <= c && x ^ y> = number c is then cut with all program 
details, see Code

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int T,A,B,C;
int a[35],b[35],c[35];
ll f[35][2][2][2][2];
void cal(int x,int v[])
{
    for (int i=0;i<=30;i++) v[i]=(x>>i)&1;
}
ll dfs(int pos,boolLima, BOOL limb of, BOOL limand, BOOL limxor)
 //           bit <a y<b x&y<cx^y> C X 
{
     IF (POS == - . 1 ) return  . 1 ;
     IF (F [POS] [Lima] [limb of] [limand] [limxor] = -! 1 ) return f [POS] [lima] [limb] [limand] [limxor]; 

    int AA = lima A [POS]:? 1 ;    // been equal before lima Description = 1 to take this bit <= a, otherwise described before <a disorder can take the current position 
    int BB = limb of B [POS]:? . 1 ;
     int C1 limand = C [POS]:? . 1 ; //Similarly, if before x & y has been equal to c then the current position to x & y <= c, otherwise you can get messy 
    int c2 = limxor c [POS]:? 0 ; 
    LL & RET = f [POS] [Lima] [limb] [limand] [limxor]; 
    RET = 0 ; 

    for ( int I = 0 ; I <= AA; I ++ )
     for ( int J = 0 ; J <= BB; J ++ ) 
    { 
        IF ((I & J)> C1) Continue ;
         IF (( I ^ J) <C2) Continue ; 
        RET + = DFS (POS- . 1 , Lima && I == AA, limb of && J == BB, limand && (I & J) == C1, limxor && (I ^ J) == C2);
    }
    return ret;
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&A,&B,&C);
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        memset(f,-1,sizeof(f));
        cal(A,a); cal(B,b); cal(C,c);
        LL ANS = DFS ( 30 , . 1 , . 1 , . 1 , . 1 ); 
        ANS - = max ( 0 , A-C + . 1 ); 
        ANS - = max ( 0 , B-C + . 1 );   // subtract x, y is 0 
        the printf ( " % LLD \ n- " , (LL) a * the B- ANS); 

    } 
    return  0 ; 
}

View Code