标签:ase 十进制 tar test pat ble file arch stream
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.
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
2
1 ab 1 2
Impossible
#include <iostream>
#include <cstring>
using namespace std;
#define Max 3
// 转成十进制
long long toDecimal(char *N,long long radix){
long long decimal=0;
for(int i=0;i<strlen(N);i++){
if(N[i]<58)
decimal=decimal*radix+(N[i]-48); // 0-9
else
decimal=decimal*radix+(N[i]-87); // a-z ,10-35 呵呵
}
return decimal;
}
//找出串中最大值
int maxValueStr(char *N){
int max=0;
for(int i=0;i<strlen(N);i++)
if(N[i]<58&&N[i]-48>max)
max=N[i]-48;
else if(N[i]-87>max)
max=N[i]-87;
if(max>1)
return max;
else
return 1;
}
int compare(char *N,long long radix,long long target){
long long decimal=0;
for(int i=0;i<strlen(N);i++){
if(N[i]<58)
decimal=decimal*radix+(N[i]-48);
else
decimal=decimal*radix+(N[i]-87);
if(decimal>target||decimal<0)
return 1;
}
if(decimal>target)
return 1;
else if(decimal<target)
return -1;
else
return 0;
}
long long binarySearch(long long target,char *N,long long low,long long high){
long long mid=low;
while(low<=high){
if(compare(N,mid,target)==1)
high=mid-1;
else if(compare(N,mid,target)==-1)
low=mid+1;
else
return mid;
mid=(low+high)/2;
}
return -1;
}
int main(int argc, char* argv[])
{
char *N[Max];
long long radix,target;
int tag;
for(int i=0;i<Max;i++)
N[i]=new char[10]();
cin >> N[1] >> N[2] >> tag >> radix;
target=toDecimal(N[tag],radix); //目标数转成十进制比较
tag=(tag==1)? 2:1;
long long max=(long long)maxValueStr(N[tag])+1;
if(target<=max){
long long r=binarySearch(target,N[tag],max,36);
if(r==-1)
cout<<"Impossible";
else
cout<<r;
}else{
long long r=binarySearch(target,N[tag],max,target+1);
if(r==-1)
cout<<"Impossible";
else
cout<<r;
}
return 0;
}
标签:ase 十进制 tar test pat ble file arch stream
原文地址:https://www.cnblogs.com/chengdalei/p/10739713.html