B

https://vjudge.net/contest/387998#problem/B

Little Rabbit is interested in radix. In a positional numeral system, the radix is the number of unique digits, including the digit $0$, used to represent numbers. For example, for the decimal system (the most common system in use today) the radix is ten, because it uses the ten digits from $0$ to $9$. Generally, in a system with radix $b$ ($b > 1$), a string of digits $d_1 dots d_n$ denotes the number $d_1b^{n-1} + d_2b^{n-2} + dots + d_nb^0$, where $0 le d_i < b$.

Little Rabbit casually writes down an equation. He wonders which radix this equation fits.

InputThe are several test cases. Each test case contains a string in a line, which represents the equation Little Rabbit writes down. The length of the string is at most $15$. The input is terminated by the end-of-file.

The equation's format: number, operator, number, $=$, number. There's no whitespace in the string.

Each number has at least $1$ digit, which may contain digital numbers $0$ to $9$ or uppercase letters $A$ to $F$ (which represent decimal $10$ to $15$). The number is guaranteed to be a non-negative integer, which means it doesn't contain the radix point or negative sign. But the number may contain leading zeros.

The operator refers to one of $+$, $-$, $*$, or $/$. It is guaranteed that the number after $/$ will not be equal to $0$. Please note that the division here is not integer division, so $7/2=3$ is not correct.OutputFor each test case, output an integer $r$ ($2 le r le 16$) in a line, which means the equation is correct in the system with radix $r$. If there are multiple answers, output the minimum one. If there is no answer between $2$ and $16$, output $-1$.Sample Input

1+1=10
18-9=9
AA*AA=70E4
7/2=3

Sample Output

2
10
16
-1

题意：
　　给一些算式 判断该算式成立的数的最小进制

　　保证都是整除，不是的话输出-1，长度为15，进制为2~16

思路：

　　模拟 分字符串

　　　字符串通过+， -， *把数字分割出来，产生3段

　　　从小到大枚举每种进制

代码：

　　

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
const int N=1e6+10;
const int mod=1e9+7;

int Atoi(string s,int radix)  
{
	int ans = 0;
	for(int i = 0;i<s.size();i++)
	{
		char t = s[i];
		if(t >= '0' && t <= '9') 
		{
			if(t - '0' >= radix)
				return -1;
			ans = ans * radix + 1ll * (t -'0');
		}
		else 
		{
			if(1ll * (t - 'A' + 10) >= radix) 
				return -1;
			ans = ans * radix + 1ll * (t - 'A' + 10);
		}
	}
	return ans;
}
signed main()
{
	string s;
	while(cin >> s)
	{
		int opt = -1;
		int eq = -1;
		for(int i = 0; i < s.size(); i++)
		{
			if(s[i] >= '0' && s[i] <= '9' || s[i] >= 'A' && s[i] <= 'Z')
			{
				continue;
			}
			if(opt == -1)
			{
				opt = i;
			}
			else
			{
				eq = i;
			}
		}
		string A = s.substr(0, opt);
		string B = s.substr(opt + 1, eq - opt - 1);
		string C = s.substr(eq + 1);
		bool flag = 0;
		for(int i = 2; i <= 16; i++)
		{
			int a = Atoi(A, i), b = Atoi(B, i), c= Atoi(C, i);
			if(a == -1 || b == -1 || c == -1)
			{
				continue;
			}
			bool S = 0;
			if(opt == '+')
			{
				S = a + b == c;
			}
			else if(s[opt] == '-')
			{
				S = a - b == c;
			}
			else if(s[opt] == '*')
			{
				S = a * b == c;
			}
			else 
			{
				if(b == 0) return -1;
				S = a % b == 0 && a == b * c;
			}
			if(S)
			{
				cout << i << endl;
				flag = 1;
				break;
			}
		}
		if(!flag)
		{
			cout << -1 << endl;
		}
	}
}

　　

 参考：https://www.cnblogs.com/lipoicyclic/p/13448226.html