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
转载来自大佬:https://blog.csdn.net/CV_Jason/article/details/80993283
代码：
#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;
} 我自己的写的代码不知道怎么说...... 是不对
代码:
#include <iostream>
#include <cctype>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
vector<long long>vec;
//转化成十进制的ll
ll Strtonum(string s, ll radix)
{
ll sum = 0;
ll radix_num = 1;
string t = s;
reverse(t.begin(),t.end());//反转下
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);//求出最大的那个
}
return maxn;
}
int main()
{
string n1,n2;
ll tag,radix,maxn1,maxn2;
cin>>n1>>n2>>tag>>radix;
//进制有个条件就是......??每个数字都不能超过....该进制
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; //假装mid就是进制
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;
}