mirror of
https://github.com/mx42/codingame.git
synced 2026-01-13 21:29:51 +01:00
132 lines
3.5 KiB
C++
132 lines
3.5 KiB
C++
// https://www.codingame.com/ide/puzzle/mayan-calculation
|
|
// Xavier Morel - 2016-03-15
|
|
|
|
#include <iostream>
|
|
#include <string>
|
|
#include <map>
|
|
#include <algorithm>
|
|
|
|
using namespace std;
|
|
|
|
/**
|
|
* Auto-generated code below aims at helping you parse
|
|
* the standard input according to the problem statement.
|
|
**/
|
|
int main()
|
|
{
|
|
int L;
|
|
int H;
|
|
cin >> L >> H; cin.ignore();
|
|
map<int, string> nums;
|
|
for (int i = 0; i < H; i++) {
|
|
string numeral;
|
|
cin >> numeral; cin.ignore();
|
|
for (int j = 0; j < 20; j++) {
|
|
if (!nums.count(j)) {
|
|
nums[j] = string();
|
|
}
|
|
nums[j] += numeral.substr(j*L, L);
|
|
nums[j] += "\n";
|
|
}
|
|
}
|
|
map<int, string> term1;
|
|
int S1;
|
|
cin >> S1; cin.ignore();
|
|
for (int i = 0; i < S1; i++) {
|
|
int numPow = floor(i/H);
|
|
string num1Line;
|
|
cin >> num1Line; cin.ignore();
|
|
if (!term1.count(numPow)) {
|
|
term1[numPow] = string();
|
|
}
|
|
term1[numPow] += num1Line;
|
|
term1[numPow] += "\n";
|
|
}
|
|
|
|
map<int, string> term2;
|
|
int S2;
|
|
cin >> S2; cin.ignore();
|
|
for (int i = 0; i < S2; i++) {
|
|
int numPow = floor(i/H);
|
|
string num2Line;
|
|
cin >> num2Line; cin.ignore();
|
|
if (!term2.count(numPow)) {
|
|
term2[numPow] = string();
|
|
}
|
|
term2[numPow] += num2Line;
|
|
term2[numPow] += "\n";
|
|
}
|
|
string operation;
|
|
cin >> operation; cin.ignore();
|
|
|
|
// cerr << "Numbers:" << endl;
|
|
double number1 = 0;
|
|
double number2 = 0;
|
|
|
|
for (int i = 0; i < 20; i++) {
|
|
// cerr << i << endl << nums[i] << endl;
|
|
for (int j = 0; j < term1.size(); j++) {
|
|
if (term1[j] == nums[i]) {
|
|
int pw = term1.size() - j - 1;
|
|
cerr << "Number1 += " << i << " * 20^ " << pw << endl;
|
|
number1 += i * pow(20.0, pw);
|
|
}
|
|
}
|
|
for (int j = 0; j < term2.size(); j++) {
|
|
if (term2[j] == nums[i]) {
|
|
int pw = term2.size() - j - 1;
|
|
cerr << "Number2 += " << i << " * 20^ " << pw << endl;
|
|
number2 += i * pow(20.0, pw);
|
|
}
|
|
}
|
|
}
|
|
// cerr << "term1:" << endl;
|
|
// for (int i = 0; i < term1.size(); i++) {
|
|
// cerr << term1[i] << endl;
|
|
// }
|
|
// cerr << number1 << endl;
|
|
|
|
// cerr << "term2:" << endl;
|
|
// for (int i = 0; i < term2.size(); i++) {
|
|
// cerr << term2[i] << endl;
|
|
// }
|
|
// cerr << number2 << endl;
|
|
|
|
cerr << number1 << operation << number2;
|
|
|
|
double res = 0;
|
|
switch (operation.c_str()[0]) {
|
|
case '+':
|
|
res = number1 + number2;
|
|
break;
|
|
case '-':
|
|
res = number1 - number2;
|
|
break;
|
|
case '/':
|
|
res = number1 / number2;
|
|
break;
|
|
case '*':
|
|
res = number1 * number2;
|
|
break;
|
|
}
|
|
|
|
cerr << " = " << res << endl;
|
|
|
|
bool started_to_write = false;
|
|
for (int i = 64; i > 0; i--) {
|
|
double curPow = pow(20.0, i - 1);
|
|
if (curPow <= res) {
|
|
int curNb = floor(res / curPow);
|
|
cerr << "Current pow: " << curPow << " - nb: " << curNb << " = " << curPow * curNb << " (res " << res << ")" << endl;
|
|
res -= (curNb * curPow);
|
|
cout << nums[curNb];
|
|
started_to_write = true;
|
|
} else if (started_to_write == true) {
|
|
cout << nums[0]; // endl should be included.
|
|
}
|
|
}
|
|
if (started_to_write == false) {
|
|
cout << nums[0];
|
|
}
|
|
}
|