hdu 6828 Little Rabbit‘s Equation 模拟

题目链接

题目大意

给你一个算式，求算式最小是在几进制下运算得以成立

题目思路

写一个k进制转换10进制的函数
将式子中的3个数都从k进制转化成10进制，然后运算
从最小可能的进制跑到16，若成立，把最小的结果记录下来输出即可
注意题目要求进制最小为2
0+0=0这种情况下，答案应输出2

代码

#include 

using namespace std;
typedef long long ll;
char number[20];
map<char, int> mp;

ll to_digit(string c, int k) {
    bool fu = false;
    int l = c.length();
    if (c[0] == '-') fu = 1;
    if (c[l - 1] == '.') l--;
    ll ans = 0;
    for (int i = fu; i < l; ++i) {
        ans *= k;
        ans += mp[c[i]];
    }
    return ans * (fu ? -1 : 1);
}

int main() {
    for (int i = 0; i < 10; i++) number[i] = '0' + i;
    for (int i = 0; i < 6; i++) number[i + 10] = 'A' + i;
    for (int i = 0; i < 16; i++) mp[number[i]] = i;
    string s;
    set<char> oper;
    oper.insert('+'), oper.insert('-'), oper.insert('*'), oper.insert('/');
    while (cin >> s) {
        int i = 1;
        char m = '0';
        for (; oper.count(s[i]) == 0; i++) if (mp[s[i]] > mp[m]) m = s[i];
        if (mp[s[0]] > mp[m]) m = s[0];
        string a = s.substr(0, i);
        char op = s[i++];
        int j = (s[i] != '-' ? i : i + 1);
        for (; s[j] != '='; j++) if (mp[s[j]] > mp[m]) m = s[j];
        string b = s.substr(i, (j++) - i);
        i = s[j] != '-' ? j : j + 1;
        for (; i < s.length(); i++) if (mp[s[i]] > mp[m]) m = s[i];
        string c = s.substr(j, s.length() - j + 1);
        bool flag = 0;
        int ans = -1;
        for (int k = max(mp[m] + 1, 2); k <= 16; k++) {
            ll na = to_digit(a, k), nb = to_digit(b, k), nc = to_digit(c, k);
            switch (op) {
                case '+':
                    if (na + nb == nc) flag = 1;
                    break;
                case '-'    :
                    if (na - nb == nc) flag = 1;
                    break;
                case '*':
                    if (na * nb == nc) flag = 1;
                    break;
                case '/':
                    if (na / nb == nc && na % nb == 0) flag = 1;
                    break;
            }
            if (flag) {
                ans = k;
                break;
            }
        }
        cout << ans << endl;
    }
    return 0;
}