poj 1331 Multiply

Multiply

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5179		Accepted: 2773

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

Sample Output

题意：给你三个数字p，q，r，问在哪个最小的进制下p*q=r成立；

注意：假设在k进制下p，q，r的每位上的数字应该要小于k；

#include <iostream>

#include <string.h>

using namespace std;

int change(char *a,int k){

	int len=strlen(a);

	int ans=0;

	for (int i=0;i<len;i++){

		ans = ans*k + a[i]-'0';

	}

	return ans;

}

bool isBig(char *a,int k){

	int len=strlen(a);

	for (int i=0;i<len;i++){

		if (a[i]-'0'>=k)

			return false;

	}

	return true;

}

int main(){

	char a[10],b[10],r[10];

	int t,i;

	cin>>t;

	while (t--){

		cin>>a>>b>>r;

		for (i=2;i<=16;i++){

			if (isBig(a,i)==true && isBig(b,i)==true && isBig(r,i)==true){

				int aa = change(a,i);

				int bb = change(b,i);

				int rr = change(r,i);

				long long ans = aa*bb;

				long long ans2 = rr;

				if (ans==ans2)

					break;

			}

		}

		if (i==17)

			i=0;

		cout<<i<<endl;

	}

	return 0;

}

巴特西

poj 1331 Multiply

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