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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

代码

#include 
   
    using namespace std;typedef long long ll;char number[20];map
    
      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
     
       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;}
     
    
   

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