文章目录
Author: CHEN, Yue
Organization: 浙江大学
Time Limit: 400 ms
Memory Limit: 64 MB
Code Size Limit: 16 KB
A1010 Radix (25point(s))
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
Code
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string>
using namespace std;
long long Pto10(string n, int p){ // p进制转化为10进制
long long result=0,product=1;
for(long long i=n.size()-1;i>=0;i--){
if(n[i]>='0'&&n[i]<='9') result+=(n[i]-'0')*product;
if(n[i]>='a'&&n[i]<='z') result+=(n[i]-'a'+10)*product;
product*=p;
}
return result;
}
int main(){
string n1,n2;
long long tag,radix;
cin>>n1>>n2>>tag>>radix;
if(tag==2) swap(n1,n2); // 保证n1存已知进制的数
long long num1=Pto10(n1,radix),num2,low=2,high,mid;
for(int i=0;i<n2.size();i++){ // 找到二分查找对应进制的下限
if(n2[i]>='0'&&n2[i]<='9') low=max((long long)n2[i]-'0'+1,low);
else low=max((long long)n2[i]-'a'+10+1,low);
}
high = max(low,num1); // 找到二分查找对应进制的上限
while(low<=high){ // 二分找符合的进制
mid=(low+high)/2;
num2=Pto10(n2,mid);
if(num1==num2){
printf("%d\n",mid);
return 0;
}
else if(num2>num1||num2<0) high=mid-1;
else low=mid+1;
}
printf("Impossible\n");
return 0;
}
Analysis
-已知4个数。分别为n1、n2、tag、radix。如果tag=1,则radix表示n1的进制。如果tag=2,则radix表示n2的进制。
-求另一个数是给出进制的数的什么进制。如:tag=1,则给出进制的数为n1,另一个数为n2。
-注意,最高能表示的进制不止是36进制。
-如何确定进制的上下限:
下限:n2(另一个数)的最大位的值+1。如:若n2中最大位为z,则下限为36。
上限:n1(给出进制的数)的值+1。如:n1用10进制表示的数为12345,则上限为12346。
重要参考:https://blog.csdn.net/Joyceyang_999/article/details/81908299