pku1331 multiply(use strtol 数字字符串转数字)

long int strtol ( const char * str, char ** endptr, int base ) <stdlib.h>

Parameters

str

C string containing the representation of an integral number.

endptr

Reference to an object of type char*, whose value is set by the function to the next character in str after the numerical value.
This parameter can also be a null pointer, in which case it is not used.

Return Value

If the base value is between 2 and 36, the format expected for the integral number is a succession of the valid digits and/or letters needed to represent integers of the specified radix (starting from '0' and up to 'z'/ 'Z' for radix 36). The sequence may optionally be preceded by a plus or minus sign and, if base is 16, an optional "0x" or "0X" prefix.
相关函数

atol	Convert string to long integer (function)

strtoul

Convert string to unsigned long integer (function)

strtod

Convert string to double (function)

1. /*pku1331
2.   Name: Multiply
3.   Date: 24-07-08 19:47
4.   Description: 题目给出q p r 其中p*q==r 在某种进制中,求最小的进制
5.   strtol:用stdlib头文件,avg1是数字字符串，avg2是返回的转化后的字符串，avg3是数字字符串的进制
6.     返回：十进制数 
7. */
8. #include<iostream>
9. using namespace std; 
10. #include<stdio.h>
11. #include<stdlib.h>
12. #define pr printf
13. int main(){
14.     char a[20],b[20],c[20];
15.     int t,m;
16.     int x,y,z,i;
17.     scanf("%d",&t);
18.     while(t--){
19.         scanf("%s%s%s",a,b,c);   
20.         m=0;
21.         i=0;
22.         while(a[i]!='/0'){if(m<a[i]-'0')m=a[i]-'0';i++;}
23.            i=0;
24.         while(b[i]!='/0'){if(m<b[i]-'0')m=b[i]-'0';i++;}
25.             i=0;
26.         while(c[i]!='/0'){if(m<c[i]-'0')m=c[i]-'0';i++;}
27.         while(m++<17){
28.                x=strtol(a,NULL,m);
29.                y=strtol(b,NULL,m);
30.                z=strtol(c,NULL,m);//由m进制转化成10进制 
31.                 if(z/y==x && z%y==0)break; 
32.             
33.             }
34.         if(m<17)pr("%d/n",m);else pr("0/n");    
35.     }
36. }
37. /*
38. 3
39. 6 9 42
40. 11 11 121
41. */