1010 Radix（25 分）
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.
Input Specification:
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.
Sample Input 1:
6 110 1 10
Sample Output 1:
2
Sample Input 2:
1 ab 1 2
Sample Output 2:
Impossible
轉載來自大佬:https://blog.csdn.net/CV_Jason/article/details/80993283
代碼：
#include<iostream>
#include<cctype>
#include<algorithm>
#include<cmath>
using namespace std;
long long str2num(string str,int radix){
long long sum = 0;
int index = 0;
int per_digit = 0;
for(auto t = str.rbegin();t!=str.rend();t++){
per_digit = isdigit(*t)? *t - '0':*t - 'a' + 10;
sum+=per_digit * pow(radix,index++);
}
return sum;
}
long long find_radix(string str,long long num){
long long result_radix = -1;
char it = *max_element(str.begin(),str.end());
long long low = (isdigit(it)?it - '0':it - 'a' + 10) + 1;
long long high = max(low,num);
while(low<=high){
long long mid = (low+high)/2;
long long temp = str2num(str,mid);
if(temp<0||temp>num){
high = mid - 1;
}else if(temp<num){
low = mid + 1;
}else{
result_radix = mid;
break;
}
}
return result_radix;
}
int main(){
string N1;
string N2;
int tag;
long long radix;
while(cin>>N1>>N2>>tag>>radix){
long long known_num = (tag==1?str2num(N1,radix):str2num(N2,radix));
long long result = find_radix((tag==1?N2:N1),known_num);
if(result!=-1)
cout<<result<<endl;
else
cout<<"Impossible"<<endl;
}
return 0;
} 我自己的寫的代碼不知道怎麼說...... 是不對
代碼:
#include <iostream>
#include <cctype>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
vector<long long>vec;
//轉化成十進制的ll
ll Strtonum(string s, ll radix)
{
ll sum = 0;
ll radix_num = 1;
string t = s;
reverse(t.begin(),t.end());//反轉下
for(int i = 0 ; i < s.size(); i++)
{
ll kk;
if(isdigit(t[i]))
kk = t[i] - '0';
else
kk = t[i] - 'a' + 10;
sum += radix_num * kk;
radix_num = radix_num * radix;
}
return sum;
}
ll findMaxchar(string s)
{
ll maxn = 0;
for(int i = 0 ; i < s.size(); i++)
{
ll kk;
if(isdigit(s[i]))
kk = s[i] - '0';
else
kk = s[i] - 'a' + 10;
maxn = max(kk, maxn);//求出最大的那個
}
return maxn;
}
int main()
{
string n1,n2;
ll tag,radix,maxn1,maxn2;
cin>>n1>>n2>>tag>>radix;
//進制有個條件就是......??每個數字都不能超過....該進制
if(tag == 1)
maxn1 = findMaxchar(n1);
else
maxn1 = findMaxchar(n2);
if(maxn1 >= radix)
{
printf("Impossible\n");
return 0;
}
ll num1 = tag == 1? Strtonum(n1,radix):Strtonum(n2,radix);
ll left = 2, right = 50, mid;
while(left <= right)
{
ll mid = (left + right) / 2; //假裝mid就是進制
ll num2 = tag == 1? Strtonum(n2,mid):Strtonum(n1,mid);
if(num2 == num1)
{
vec.push_back(mid);
right = mid - 1;
}else if(num2 > num1)
right = mid - 1;
else if(num2 < num1)
left = mid + 1;
}
if(tag == 1)
maxn2 = findMaxchar(n2);
else
maxn2 = findMaxchar(n1);
if(vec.size() == 0)
printf("Impossible\n");
else if(maxn2 >= vec[vec.size() - 1])
printf("Impossible\n");
else
printf("%lld\n",vec[vec.size() - 1]);
return 0;
}