Submission #875302

#	Time	Username	Problem	Language	Result	Execution time	Memory
875302		Itamar	Alice, Bob, and Circuit (APIO23_abc)	C++17	28 / 100	148 ms	16296 KiB

abc

#include "abc.h"

// you may find the definitions useful
const int OP_ZERO = 0;  // f(OP_ZERO,    x0, x1) = 0
const int OP_NOR = 1;  // f(OP_NOR,     x0, x1) = !(x0 || x1)
const int OP_GREATER = 2;  // f(OP_GREATER, x0, x1) = (x0 > x1)
const int OP_NOT_X1 = 3;  // f(OP_NOT_X1,  x0, x1) = !x1
const int OP_LESS = 4;  // f(OP_LESS,    x0, x1) = (x0 < x1)
const int OP_NOT_X0 = 5;  // f(OP_NOT_X0,  x0, x1) = !x0
const int OP_XOR = 6;  // f(OP_XOR,     x0, x1) = (x0 ^ x1)
const int OP_NAND = 7;  // f(OP_NAND,    x0, x1) = !(x0 && x1)
const int OP_AND = 8;  // f(OP_AND,     x0, x1) = (x0 && x1)
const int OP_EQUAL = 9;  // f(OP_EQUAL,   x0, x1) = (x0 == x1)
const int OP_X0 = 10; // f(OP_X0,      x0, x1) = x0
const int OP_GEQ = 11; // f(OP_GEQ,     x0, x1) = (x0 >= x1)
const int OP_X1 = 12; // f(OP_X1,      x0, x1) = x1
const int OP_LEQ = 13; // f(OP_LEQ,     x0, x1) = (x0 <= x1)
const int OP_OR = 14; // f(OP_OR,      x0, x1) = (x0 || x1)
const int OP_ONE = 15; // f(OP_ONE,     x0, x1) = 1
using namespace std;
#include <vector>
#define vi vector<int>

void mul(long long a, long long b,vi &ans) {
    long long c = (a * b);
    //c = c % (1 << 16);
    for (int i = 0; i < 16; i++) {
        ans[i] = (c % 2);
        c /= 2;
    }
}
// Alice
int // returns la
alice(
    /*  in */ const int n,
    /*  in */ const char names[][5],
    /*  in */ const unsigned short numbers[],
    /* out */ bool outputs_alice[]
) {
       for (int i = 0; i < n ; i++) {
        vi v(16);
        mul((long long)numbers[i],1,v);
        for (int j = 0; j < 16; j++) {
            outputs_alice[16 * i + j] = v[j];
        }
    }

    return 16*n;
}


// Bob
int // returns lb
bob(
    /*  in */ const int m,
    /*  in */ const char senders[][5],
    /*  in */ const char recipients[][5],
    /* out */ bool outputs_bob[]
) {
    for (int i = 0; i < 26 * 26; i++)outputs_bob[i] = 0;
    for (int i = 0; i < m; i++) {
        outputs_bob[(26 * (recipients[i][0] - 'a')) + senders[i][0] - 'a'] = 1;
    }
    return 26*26;
}


// Circuit
int // returns l
circuit(
    /*  in */ const int la,
    /*  in */ const int lb,
    /* out */ int operations[],
    /* out */ int operands[][2],
    /* out */ int outputs_circuit[][16]
) {
    // and = 8, xor= 6, or = 14,x1=12;
    int last = la + lb;
    for (int i = 0; i < 26; i++) {
        for (int k = 0; k < 16; k++) {
            operands[last + k][0] = 0;
            operands[last + k][1] = 0;
            operations[last + k] = 0;

        }
        last += 16;
        for (int j = 0; j < 26; j++) {
            for (int k = last; k < last + 16; k++) {
                operations[k] = OP_AND;
                operands[k][0] = la+(26 * i) + j;
                operands[k][1] = 16 * j + (k-last);
            }
            // epsilon_i,j*a_i=operands[last,last+1,...last+15]
            for (int k = 0; k < 16; k++) {

                    operations[last + 6*k+16] = OP_XOR;
                    operands[last + 6*k+16][0] = last-16+k;
                    operands[last + 6*k + 16][1] = last + k;

                    if (k>0)operations[last + 6 * k + 17] = OP_XOR; else operations[last + 6 * k + 17] = OP_X0;
                        operands[last + 6 * k + 17][0] = last + 16 + (6 * k);
                        operands[last + 6 * k + 17][1] = last + 6 * k + 15;

                
                    operations[last + 6 * k + 18] = OP_OR; 
                    operands[last + 6 * k + 18][0] = last-16+k;
                    operands[last + 6 * k + 18][1] = last + k;

                    if (k)operations[last + 6 * k + 19] = OP_AND; else operations[last + 6 * k + 19] = 0;
                    operands[last + 6 * k + 19][0] = last + 6 * k + 18;
                    operands[last + 6 * k + 19][1] = last + 6 * k + 15;

                    operations[last + 6 * k + 20] = OP_AND;
                    operands[last + 6 * k + 20][0] = last  -16+k;
                    operands[last + 6 * k + 20][1] = last + k;

                    operations[last + 6 * k + 21] = OP_OR;
                    operands[last + 6 * k + 21][0] = last + 6 * k + 20;
                    operands[last + 6 * k + 21][1] = last + 6 * k + 19;
                    
            }
            for (int k = last + 112; k < last + 112+16; k++) {
                operations[k] = OP_X1;
                operands[k][0] = 0;
                operands[k][1] = 6*(k - last - 112) + last + 17;
            }
            last += 128;
        }
        for (int k = 0; k < 16; k++) {
            outputs_circuit[i][k] = last-16 + k;
        }
    }

    return last;
}

• Subtask 1: 4 / 4
• Subtask 2: 0 / 4
• Subtask 3: 0 / 4
• Subtask 4: 24 / 24
• Subtask 5: 0 / 24
• Subtask 6: 0 / 6
• Subtask 7: 0 / 18
• Subtask 8: 0 / 10
• Subtask 9: 0 / 6