1010 Radix
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
代码
自己代码22分
#include <iostream>
using namespace std;
long long getNum(string str,long long radix){
//作用：按照基数radix，将字符串转化str为数
//可能会很大，用longlong
long long sum=0;
for (int i = 0; i < str.length(); ++i) {
if (str[i] <= '9') {
sum = sum * radix + (str[i] - '0');
}
else if(str[i]<='z') {
sum = sum*radix+(str[i]-'a'+10);
}
}
return sum;
}
int main() {
string str1,str2;
int tag;
long long radix,num1;
cin >> str1 >> str2 >> tag >> radix;
if(tag == 1){
num1=getNum(str1,radix);
}
else{
// 如果tag=2，则将str1和str2交换
num1 = getNum(str2,radix);
str2 = str1;
}
bool flag = false;
for(int i=2; i<num1; i++) {
if(getNum(str2,i) == num1) {
cout << i;
flag = true;
break;
}
}
if(!flag){
cout << "Impossible";
}
return 0;
}
满分代码
#include <iostream>
using namespace std;
long long getNum(string str,long long radix){
//作用：按照基数radix，将字符串转化str为数
//可能会很大，用longlong
long long sum=0;
for (int i = 0; i < str.length(); ++i) {
if (str[i] <= '9') {
sum = sum * radix + (str[i] - '0');
}
else if(str[i]<='z') {
sum = sum*radix+(str[i]-'a'+10);
}
}
return sum;
}
long long getRadix(long long num1,string str,long long start,long long end){
if(start==end){
//等号单独讨论，否则可能死循环
if(getNum(str,start)==num1) {
return start;
}
else{
return 0;
}
}
else if(start<end){
long long radix=(start+end)/2;
//二分查找
if(getNum(str,radix) == num1) {
return radix;
}else if(getNum(str,radix)>num1 || getNum(str,radix)<0) {
//num2>num1或num2<0都表示这个radix取得过大了，需要减小
return getRadix(num1,str,start,radix);
}else {
//num2<num1,说明radix取得比较小，所以要往大了取
return getRadix(num1,str,radix+1,end);
}
}
return 0;
}
int getMinRadix(string num2){
//作用：找到num2的最小基数
//例如：num2中最大的权值为b时，最小基数就是b+1=12
char max='0';
for (int i = 0; i < num2.length(); ++i) {
if(num2[i] > max) {
max = num2[i];
}
}
if (max <= '9') {
return max-'0'+1;
}
return max-'a'+11;
}
int main() {
string str1,str2;
int tag;
long long radix,num1;
cin >> str1 >> str2 >> tag >> radix;
if(tag == 1){
num1=getNum(str1,radix);
}
else{
// 如果tag=2，则将str1和str2交换
num1 = getNum(str2,radix);
str2 = str1;
}
int res = getRadix(num1, str2, getMinRadix(str2), num1+1);
if(res == 0){
cout << "Impossible";
return 0;
}
cout << res;
return 0;
}