poj 1331 Multiply

Multiply

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4634		Accepted: 2466

Description

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.

Input

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.

Output

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.

Sample Input

3
6 9 42
11 11 121
2 2 2

Sample Output

13
3
0

#include<iostream>
using namespace std;

char q[10];
char p[10];
char r[10];
int maxn;

void find_max()
{
int i;
    maxn=0;
for(i=0;i<strlen(q);i++)
    {
if(maxn<(q[i]-'0'))
            maxn=(q[i]-'0');
    }
for(i=0;i<strlen(p);i++)
    {
if(maxn<(p[i]-'0'))
            maxn=(p[i]-'0');
    }
for(i=0;i<strlen(r);i++)
    {
if(maxn<(r[i]-'0'))
            maxn=(r[i]-'0');
    }
}

int change2num(char *p,int len,int b)
{
int i;
int sum=0;
int weight=1;
for(i=len-1;i>=0;i--)
    {
        sum+=(p[i]-'0')*weight;
        weight*=b;
    }
return sum;
}

int main()
{
int n;
int i;
int nump, numq, numr;
int multi;
    cin>>n;
while(n--)
    {
        cin>>p;
        cin>>q;
        cin>>r;
        find_max();
for(i=maxn+1;i<=16;i++)
        {
            nump=change2num(p,strlen(p),i);
            numq=change2num(q,strlen(q),i);
            numr=change2num(r,strlen(r),i);
            multi=nump*numq;
if(multi==numr)
            {
                cout<<i<<endl;
break;
            }
        }
if(i==17)
            cout<<"0"<<endl;
    }
return 0;
}

转载于:https://www.cnblogs.com/w0w0/archive/2011/11/21/2256994.html