Multiply

Multiply

总时间限制: 1000ms 内存限制: 65536kB
描述
6*9 = 42” is not true for base 10, but is true for base 13. That is, 6(13) * 9(13) = 42(13) because 42(13) = 4 * 131 + 2 * 130 = 54(10).

You are to write a program which inputs three integers p, q, and r and determines the base B (2<=B<=16) for which p * q = r. If there are many candidates for B, output the smallest one. For example, let p = 11, q = 11, and r = 121. Then we have 11(3) * 11(3) = 121(3) because 11(3) = 1 * 31 + 1 * 30 = 4(10) and 121(3) = 1 * 32 + 2 * 31 + 1 * 30 = 16(10). For another base such as 10, we also have 11(10) * 11(10) = 121(10). In this case, your program should output 3 which is the smallest base. If there is no candidate for B, output 0.
输入
The input consists of T test cases. The number of test cases (T ) is given in the first line of the input file. Each test case consists of three integers p, q, and r in a line. All digits of p, q, and r are numeric digits and 1<=p,q, r<=1,000,000.
输出
Print exactly one line for each test case. The line should contain one integer which is the smallest base for which p * q = r. If there is no such base, your program should output 0.
样例输入
3
6 9 42
11 11 121
2 2 2
样例输出
13
3
0

#include <iostream>
#include<string.h>
using namespace std;
//http://bailian.openjudge.cn/exam/3658/3/
//b进制转10进制，一般都是用char数组存储的 
int n,flag;
char p[8],q[8],r[8];
int pp,qq,rr;
int b2t(char x[],int b){
    int res=0,len=strlen(x);
    for(int i=0;i<len;i++){
        if(x[i]-'0'>=b){
            return -1;
        }
        else{
            res=res*b+(x[i]-'0');
        }
    }
    return res;
}
int main(int argc, char *argv[]) {
    cin>>n;
    while(n--){
        flag=0;
        cin>>p>>q>>r;
        for(int i=2;i<=16;i++){
            pp=b2t(p,i);
            if(pp!=-1){
                qq=b2t(q,i);
                if(qq!=-1){
                    rr=b2t(r,i);
                    if(rr!=-1&&rr==pp*qq){
                        cout<<i<<endl;
                        flag=1;
                        break;
                    }
                }
            }
        }
        if(flag==0){
            cout<<"0"<<endl;
        }
    }
    return 0;
}