Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is yes, if 6 is a decimal number and 110 is a binary number.
Now for any pair of positive integers N1 and N2, your task is to find the radix of one number while that of the other is given.
Each input file contains one test case. Each case occupies a line which contains 4 positive integers:
N1 N2 tag radix
Here N1 and N2 each has no more than 10 digits. A digit is less than its radix and is chosen from the set { 0-9, a-z } where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The last number radix is the radix of N1 if tag is 1, or of N2 if tag is 2.
Output Specification:
For each test case, print in one line the radix of the other number so that the equation N1 = N2 is true. If the equation is impossible, print Impossible. If the solution is not unique, output the smallest possible radix.
6 110 1 10
Sample Output 1:
2
1 ab 1 2
Sample Output 2:
Impossibl数学题一定要考虑溢出的情况
1 #include<iostream>
2 #include<
string>
3 #include<cmath>
4 using namespace std;
5 /*
6 烦人的一题， 难点在于数字溢出！
7 还有就是选取最小的进制， 找出字符串中最大的字母，则最小进制为最大字母代表的数字+1
8 如果最小进制的值比num的值大，那么最大进制也为最小进制
9 在进行数的运算的时候 要考虑数字溢出的情况 当整数溢出的时候， 结果会小于0
10 此外进制的上限并非36
11 */
12 long long convert(
string s,
long long radix){
13 long long ans=
0;
14 for(
int i=
0; i<s.size(); i++
){
15 long long temp = (s[i]>=
'0' && s[i]<=
'9') ? (s[i]-
'0') : (s[i]-
'a'+
10);
16 ans = ans*radix +
temp;
17 }
18 return ans;
19 }
20 const long long inf = pow(
2,
62)-
1;
21 long long binary(
string& s,
long long num){
22 char maxx=
'0';
23 for(
int i=
0; i<s.size(); i++
)
24 if(s[i]>maxx) maxx=
s[i];
25 long long low=(maxx<=
'9' && maxx>=
'0' ? maxx-
'0' : maxx-
'a'+
10)+
1, high = low<num?
num :low;
26 while(low<=
high){
27 long long mid=(low+high)/
2, number=
convert(s, mid);
28 if(number>num || number<
0) high=mid-
1;
///
29 else if(number<num) low=mid+
1;
30 else return mid;
31 }
32 return -
1;
33 }
34 int main(){
35 string num1, num2;
36 long long radix, tag, num;
37 cin>>num1>>num2>>tag>>
radix;
38 if(tag==
1){
39 num =
convert(num1, radix);
40 radix=
binary(num2, num);
41 }
else{
42 num =
convert(num2, radix);
43 radix=
binary(num1, num);
44 }
45 if(radix==-
1) cout<<
"Impossible"<<
endl;
46 else cout<<radix<<
endl;
47 return 0;
48 }
转载于:https://www.cnblogs.com/mr-stn/p/9572736.html
