POJ 1331 进制转换

寻找特殊基数解

最新推荐文章于 2021-05-23 23:04:04 发布

最新推荐文章于 2021-05-23 23:04:04 发布 · 98 阅读

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#c/c++

本文介绍了一个编程问题，即寻找使得两个数相乘等于第三个数的最小基数B（2<=B<=16）。通过输入三个整数p、q和r，程序需要找出满足p*q=r的最小基数B，若无解则输出0。示例展示了如何处理多个测试用例。

Multiply

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4287		Accepted: 2250

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

13
3
0

Source

Taejon 2002

Source Code

Problem: 1331		User: bingshen
Memory: 144K		Time: 0MS
Language: C++		Result: Accepted

Source Code

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int main()
{
	char a[100],b[100],c[100];
	int i,x,y,z,t,l1,l2,l3,m1,m2,m3,m;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s%s%s",a,b,c);
		m=0;
		bool flag=false;
		l1=strlen(a);
		l2=strlen(b);
		l3=strlen(c);
		for(i=0;i<l1;i++)
		{
			m1=a[i]-'0';
			if(m1>m)
				m=m1;
		}
		for(i=0;i<l2;i++)
		{
			m2=b[i]-'0';
			if(m2>m)
				m=m2;
		}
		for(i=0;i<l3;i++)
		{
			m3=c[i]-'0';
			if(m3>m)
				m=m3;
		}
		for(i=m+1;i<=16;i++)
		{
			x=strtol(a,NULL,i);
			y=strtol(b,NULL,i);
			z=strtol(c,NULL,i);
			if(x*y==z)
			{
				flag=true;
				printf("%d/n",i);
				break;
			}
		}
		if(!flag)
			printf("0/n");
	}
	return 0;
}