Algorithm
Problem Name: 2 AD-HOC - beecrowd | 1945
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/1945
Simulator
By Emilio Wuerges, UFFS 
Brazil
Timelimit: 1
Centuries after the alien invasion, after mankind had been rebuild completely, a very old set of programs was discovered in an obsolete programming language, called Java++. For historical curiosity, you were assigned to understand the meaning of these programs.
Your task is to write a simulator for these programs, and as initial test, this simulator must be able to calculate the value of the last assigned variable of each program.
Input
Each input consists of 1 program. The program only contains 2 kinds of instructions. One instruction to assign a variable and one to perform a sum.
Assignment instructions have the format:
A := B
Where A is a variable name and B is a positive integer.
Sum instructions have the format:
A := B + C
Where A is a variable name and B or C are either a variable name or a positive integer.
Tokens of these programs are always separated by spaces and instructions are separated by a line break.
Only combinations of at most 8 lower case letters are considered as valid variable names.
Programs have, at most, 2000 instructions
Variables are assigned, at most, only once.
Output
Output consists of only 1 line, with only one positive integer number, containing the last value of the last variable assigned, be it by a direct assignment or by a sum.
| Input Sample | Output Sample |
|
a := 2 |
3 |
| Input Sample | Output Sample |
|
a := 0 |
7 |
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
map < string, long long> m;
long long toint(string &a)
{
long long ret = 0;
for (int i = 0 ; i < a.size(); ++i)
{
ret *= 10LL;
ret += (long long)(a[i] - '0');
}
return ret;
}
int main()
{
ios_base :: sync_with_stdio(0); cin.tie(0);
string in;
string salva;
stringstream ss;
string aux, aux2, aux3, op1, op2;
while (getline(cin, in))
{
salva = in;
ss << in;
ss >> aux;
ss >> op1;
ss >> aux2;
if (ss >> op2)
{
ss >> aux3;
if (isdigit(aux2[0]))
{
if (isdigit(aux3[0]))
{
long long a = toint(aux2), b = toint(aux3);
m[aux] = a + b;
}
else
{
long long a = toint(aux2);
m[aux] = a + m[aux3];
}
}
else
{
if (isdigit(aux3[0]))
{
long long b = toint(aux3);
m[aux] = m[aux2] + b;
}
else
m[aux] = m[aux2] + m[aux3];
}
}
else
m[aux] = toint(aux2);
ss.clear();
}
ss << salva;
ss >> aux;
cout << m[aux] << '\n';
m.clear();
ss.clear();
}
Input
b := 1
c := a + b
d := a + b
Output