Educational Codeforces Round 66 (Rated for Div. 2) B. Catch Overflow!

链接：https://codeforces.com/contest/1175/problem/B

题意：

You are given a function ff written in some basic language. The function accepts an integer value, which is immediately written into some variable xx. xx is an integer variable and can be assigned values from 00 to 232−1232−1. The function contains three types of commands:

• for nn — for loop;
• end — every command between "for nn" and corresponding "end" is executed nn times;
• add — adds 1 to xx.

After the execution of these commands, value of xx is returned.

Every "for nn" is matched with "end", thus the function is guaranteed to be valid. "for nn" can be immediately followed by "end"."add" command can be outside of any for loops.

Notice that "add" commands might overflow the value of xx! It means that the value of xxbecomes greater than 232−1232−1 after some "add" command.

Now you run f(0)f(0) and wonder if the resulting value of xx is correct or some overflow made it incorrect.

If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of xx.

思路：

模拟，我用递归写的。

用stack也可以

代码：

#include <bits/stdc++.h>
using namespace std;
 
typedef long long LL;
const unsigned int MAXV = (1LL<<32)-1;
const int MAXN = 1e5+10;
int n;
bool flag = true;
 
struct Op
{
    string op;
    LL va;
}oper[MAXN];
int step = 1;
 
LL Dfs(LL lops)
{
    if (step > n)
        return 0;
    LL res = 0;
    while (oper[step].op[0] == 'a')
    {
        step++;
        res++;
    }
//    cout << "res1:" << res << endl;
    while (oper[step].op[0] == 'f')
    {
        res += Dfs(oper[step++].va);
        if (res > MAXV)
            flag = false;
        while (oper[step].op[0] == 'a')
        {
            step++;
            res++;
        }
    }
//    cout << "res2:" << res << endl;
    if (1LL*lops*res > MAXV)
        flag = false;
//    cout << "res3:" << lops*res << ' ' << step << endl;
    step++;
    return lops*res;
}
 
int main()
{
//    freopen("test.in", "r", stdin);
    cin >> n;
    for (int i = 1;i <= n;i++)
    {
        cin >> oper[i].op;
        if (oper[i].op[0] == 'f')
            cin >> oper[i].va;
    }
    LL res = Dfs(1);
    if (!flag)
        cout << "OVERFLOW!!!" << endl;
    else
        cout << res << endl;
 
    return 0;
}