老手的解法

#include <iostream>
#include <climits>
using namespace std;

int main() {
    int xl, xu, yl, yu;
    cin >> xl >> xu >> yl >> yu;

    // 检查所有可能的极值组合是否可能溢出
    bool overflow = false;
    if ((long long)xl * yl < INT_MIN || (long long)xl * yl > INT_MAX) overflow = true;
    if ((long long)xl * yu < INT_MIN || (long long)xl * yu > INT_MAX) overflow = true;
    if ((long long)xu * yl < INT_MIN || (long long)xu * yl > INT_MAX) overflow = true;
    if ((long long)xu * yu < INT_MIN || (long long)xu * yu > INT_MAX) overflow = true;

    cout << (overflow ? "long long int" : "int") << endl;
    return 0;
}

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