Multiply
Time Limit: 1000MS | Memory Limit: 10000K |
Total Submissions: 5943 | Accepted: 3084 |
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
Source
Taejon 2002
Regionals 2002 >> Asia - Taejon
问题链接:POJ1331 UVALive2526 Multiply
问题简述:(略)
问题分析:
这是一个简单的进制转换问题。
输入的数据都是有数字(0-9)组成的。
由于如果有两个答案的话则输出小的值,所以需要从小到大试探进制。
开始时,先将数据按照10进制读入,然后再转换为指定的进制。
程序说明:
编写了共用的函数convert()用于转换数据。
题记:
将共用功能用封装到函数中是一种好的做法。
参考链接:(略)
AC的C语言程序如下:
/* POJ1331 UVALive2526 Multiply */
#include <stdio.h>
#define BASE10 10
#define START 2
#define END 16
int convert(int val, int base)
{
int ans, weight, r;
ans = 0;
weight = 1;
while(val) {
r = val % BASE10;
val /= BASE10;
if(r >= base)
return -1;
ans += weight * r;
weight *= base;
}
return ans;
}
int main(void)
{
int t, p, q, r, i;
int p2, q2, r2;
scanf("%d", &t);
while(t--) {
scanf("%d%d%d", &p, &q, &r);
for(i=START; i<=END; i++) {
p2 = convert(p, i);
if(p2 < 0)
continue;
q2 = convert(q, i);
if(q2 < 0)
continue;
r2 = convert(r, i);
if(r2 < 0)
continue;
if(p2 * q2 == r2)
break;
}
printf("%d\n", i <= END ? i : 0);
}
return 0;
}