POJ 1331 Mutiply strtol()进制转换函数

http://poj.org/problem?id=1331

 

Multiply

Time Limit: 1000MS

Memory Limit: 10000K

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

 

函数声明

 

函数及其参数

　　long int strtol(const char *nptr,char **endptr,int base);

函数的解释说明：

　　这个函数会将参数nptr字符串根据参数base来转换成长整型数。参数base范围从2至36，或0。参数base代表采用的进制方式，如base值为10则采用10进制，若base值为16则采用16进制等。当base值为0时则是采用10进制做转换，但遇到如’0x’前置字符则会使用16进制做转换、遇到’0’前置字符而不是’0x’的时候会使用8进制做转换。一开始strtol()会扫描参数nptr字符串，跳过前面的空格字符，直到遇上数字或正负符号才开始做转换，再遇到非数字或字符串结束时('/0')结束转换，并将结果返回。若参数endptr不为NULL，则会将遇到不合条件而终止的nptr中的字符指针由endptr返回。

　　strtol是atoi的增强版，主要体现在这几方面：

　　1.不仅可以识别十进制整数，还可以识别其它进制的整数，取决于base参数，比如strtol("0XDEADbeE~~", NULL, 16)返回0xdeadbee的值，strtol("0777~~", NULL, 8)返回0777的值。

　　2.endptr是一个传出参数，函数返回时指向后面未被识别的第一个字符。例如char *pos; strtol("123abc", &pos, 10);，strtol返回123，pos指向字符串中的字母a。如果字符串开头没有可识别的整数，例如char *pos; strtol("ABCabc", &pos, 10);，则strtol返回0，pos指向字符串开头，可以据此判断这种出错的情况，而这是atoi处理不了的。

 

　　3.如果字符串中的整数值超出long int的表示范围（上溢或下溢），则strtol返回它所能表示的最大（或最小）整数，并设置errno为ERANGE，例如strtol("0XDEADbeef~~", NULL, 16)返回0x7fffffff并设置errno为ERANGE

/* Author : yan * Question : POJ 1331 Multiply * Date && Time : Friday, February 11 2011 05:54 PM * Compiler : gcc (Ubuntu 4.4.3-4ubuntu5) 4.4.3 */ #include<stdio.h> char mul1[10],mul2[10],res[10]; int main() { //freopen("input","r",stdin); int n; int base; int m1,m2,m3; scanf("%d",&n); while(n--) { scanf("%s %s %s",mul1,mul2,res); base=2; while(base<17) { m1=strtol( mul1,'/0',base ); m2=strtol( mul2,'/0',base ); m3=strtol( res,'/0',base ); if(m1*m2==m3 && m1!=0 && m2!=0) { printf("%d/n",base); goto end; } base++; } if(base==17) printf("0/n"); end:; } return 0; }