2019快手秋招，字符串长度最大乘积

暴力解法
时间复杂度：O(n^2)
空间复杂度：O(n)
测试用例有问题，AC=75%??? are you sure?

#include <iostream>
#include <vector>
#include <string>

using namespace std;
//判断两个字符串是否为包含关系
bool isIncludeString(string s0, string s1) {
    int len0 = s0.size();
    int len1 = s1.size();
    bool flag;
    for (int i = 0; i <= len0; i++) {
        for (int j = 0; j < len1; j++) {
            if (s0[i] == s1[j]) {
                return true;
            }
        }
    }
    return false;
}

int main() {
    vector<string> vec;
    string s;
    string c;
    while (cin >> s) {
        if (s == "end")
        {
            break;
        }
        vec.push_back(s);
    }
    int res = 0;
    int maxn = 0;
    for (int i = 0; i < vec.size() - 1; i++) {
        for (int j = i + 1; j < vec.size(); j++) {
            if (isIncludeString(vec[i], vec[j])) {
                continue;
            }
            else {
                //更新最大值
                maxn = max(res, (int)(vec[i].size() * vec[j].size()));
            }
        }
    }
    cout << maxn << endl;
    return 0;
}