Submission #768180

#	Time	Username	Problem	Language	Result	Execution time	Memory
768180		green_gold_dog	Alice, Bob, and Circuit (APIO23_abc)	C++17	66 / 100	435 ms	494880 KiB

abc

#include "abc.h"
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

// 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

constexpr ll LOG = 16, COLC = 27, ADD = 1, LOG2 = 20, SLOGA = LOG + LOG2, SLOGB = LOG2 * 2;

// Alice
int // returns la
alice(
    /*  in */ const int n,
    /*  in */ const char name[][5],
    /*  in */ const unsigned short num[],
    /* out */ bool oa[]
) {
	ll last = 0;
	for (ll i = 0; i < n; i++) {
		ll now = 0;
		for (auto j : name[i]) {
			if (j < 'a') {
				break;
			}
			now *= COLC;
			now += j - 'a' + ADD;
		}
		for (ll j = 0; j < LOG2; j++) {
			oa[last] = now % 2;
			now /= 2;
			last++;
		}
		now = num[i];
		for (ll j = 0; j < LOG; j++) {
			oa[last] = now % 2;
			now /= 2;
			last++;
		}
	}
	return last;
}


// Bob
int // returns lb
bob(
    /*  in */ const int m,
    /*  in */ const char s[][5],
    /*  in */ const char r[][5],
    /* out */ bool ob[]
) {
	ll last = 0;
	for (ll i = 0; i < m; i++) {
		ll now = 0;
		for (auto j : s[i]) {
			if (j < 'a') {
				break;
			}
			now *= COLC;
			now += j - 'a' + ADD;
		}
		for (ll j = 0; j < LOG2; j++) {
			ob[last] = now % 2;
			now /= 2;
			last++;
		}
		now = 0;
		for (auto j : r[i]) {
			if (j < 'a') {
				break;
			}
			now *= COLC;
			now += j - 'a' + ADD;
		}
		for (ll j = 0; j < LOG2; j++) {
			ob[last] = now % 2;
			now /= 2;
			last++;
		}
	}
	return last;
}

pair<ll, ll> sum2(ll a, ll b, ll& last, int o[], int op[][2]) {
	o[last] = OP_XOR;
	op[last][0] = a;
	op[last][1] = b;
	last++;
	o[last] = OP_AND;
	op[last][0] = a;
	op[last][1] = b;
	last++;
	return make_pair(last - 2, last - 1);
}

pair<ll, ll> sum3(ll a, ll b, ll c, ll& last, int o[], int op[][2]) {
	auto[f, s] = sum2(a, b, last, o, op);
	auto[f2, s2] = sum2(f, c, last, o, op);
	o[last] = OP_OR;
	op[last][0] = s;
	op[last][1] = s2;
	last++;
	return make_pair(f2, last - 1);
}

vector<ll> sum(vector<ll> s1, vector<ll> s2, ll& last, int o[], int op[][2]) {
	vector<ll> ans;
	auto[f, s] = sum2(s1[0], s2[0], last, o, op);
	ans.push_back(f);
	ll per = s;
	for (ll i = 1; i < LOG; i++) {
		auto[f, s] = sum3(s1[i], s2[i], per, last, o, op);
		ans.push_back(f);
		per = s;
	}
	return ans;
}

ll check(vector<ll> fir, vector<ll> sec, ll& last, int o[], int op[][2]) {
	ll n = fir.size();
	for (ll i = 0; i < n; i++) {
		o[last] = OP_EQUAL;
		op[last][0] = fir[i];
		op[last][1] = sec[i];
		last++;
	}
	ll end = last - 1;
	for (ll i = last - n; i != end; i++) {
		o[last] = OP_AND;
		op[last][0] = i;
		op[last][1] = last - 1;
		last++;
	}
	return last - 1;
}

// Circuit
int // returns l
circuit(
    /*  in */ const int la,
    /*  in */ const int lb,
    /* out */ int o[],
    /* out */ int op[][2],
    /* out */ int oc[][16]
) {
	ll n = la / SLOGA;
	ll m = lb / SLOGB;
	if (n == 0) {
		return 0;
	}
	vector<vector<ll>> name(n), num(n), from(m), to(m);
	vector<vector<ll>> nans(n);
	ll last = 0;
	for (ll i = 0; i < n; i++) {
		for (ll j = 0; j < LOG2; j++) {
			name[i].push_back(last);
			last++;
		}
		for (ll j = 0; j < LOG; j++) {
			num[i].push_back(last);
			last++;
		}
	}
	for (ll i = 0; i < m; i++) {
		for (ll j = 0; j < LOG2; j++) {
			from[i].push_back(last);
			last++;
		}
		for (ll j = 0; j < LOG2; j++) {
			to[i].push_back(last);
			last++;
		}
	}
	ll ZERO = last;
	o[last] = OP_ZERO;
	op[last][0] = 0;
	op[last][1] = 1;
	last++;
	for (ll i = 0; i < m; i++) {
		vector<ll> sall;
		for (ll j = 0; j < n; j++) {
			ll need = check(name[j], from[i], last, o, op);
			vector<ll> nnum;
			for (ll s = 0; s < LOG; s++) {
				nnum.push_back(last);
				o[last] = OP_AND;
				op[last][0] = need;
				op[last][1] = num[j][s];
				last++;
			}
			if (sall.empty()) {
				sall = nnum;
			} else {
				vector<ll> nsall;
				for (ll s = 0; s < LOG; s++) {
					nsall.push_back(last);
					o[last] = OP_OR;
					op[last][0] = sall[s];
					op[last][1] = nnum[s];
					last++;
				}
				sall = nsall;
			}
		}
		for (ll j = 0; j < n; j++) {
			ll need = check(name[j], to[i], last, o, op);
			vector<ll> nnum;
			for (ll s = 0; s < LOG; s++) {
				nnum.push_back(last);
				o[last] = OP_AND;
				op[last][0] = need;
				op[last][1] = sall[s];
				last++;
			}
			if (nans[j].empty()) {
				nans[j] = nnum;
			} else {
				nans[j] = sum(nans[j], nnum, last, o, op);
			}
		}
	}
	for (ll i = 0; i < n; i++) {
		if (nans[i].empty()) {
			nans[i].resize(LOG, ZERO);
		}
		for (ll s = 0; s < LOG; s++) {
			oc[i][s] = nans[i][s];
		}
	}
	return last;
}

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