1010 Radix (25)（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
/********************** author: yomi date: 18.6.8 现在来梳理思路： 1. 把已知进制的数转换为相应的已给的进制 （为什么要转呢？因为他给你的这个数并不一定是正确的表示，比如说 ab 二进制 ， 那么就需要把ab先转到二进制） 2. 寻找未知进制数的进制，使他与前面的数相等--->进制数一定是从小到大遍历的，那么可以用到二分 然而还有一些小问题要去攻克： 1.二分上下界的确定：二分下界：进制必须能够表示这个数，所以该进制必须大于这个数中最大的那位数，故下界为：最大的那位数+1二分上界：分两种情况：一是已知数大于下界，那么上界为：已知数二是已知数小于下界，那么上界为：下界（容易想到的是上界为已知数，而由于上界不能小于下界，故此时上界与下界相等）这样就归结成了一个问题，为什么上界为已知数？需要指出的是 A digit is less than its radix and is chosen from the set {0-9, a-z} where 0-9represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. 只是说某一位数的进制为36，而我却以为我们要找的进制也在36进制以内，结果就WA了一半。实际上，题目并没有提上界，而当要找的进制数比已知数大的时候，这两个数就决不可能再相等了。所以将上界设置为已知数即可（一定是转换到已知进制数的已知数）。 看了大神的代码收获很多： 1. char it = *max_element(n.begin(), n.end());//查找最大元素的好办法整型数据只要在vector中就可以使用啦 2. 适当使用三目运算符确实能使代码简洁不少，这样有助于考场上分析。不管你们懂没懂，反正我懂了。 *********************分析为原创，代码部分原创 改编自柳婼******************/#include <iostream> #include <cctype> #include <algorithm> #include <cmath> #include <cstdio> using namespace std; long long convert(string n, long long radix) {
long long sum = 0;
int index = 0, temp = 0;
for(int i=0; i<n.length(); i++){
temp = isdigit(n[i]) ? n[i]-'0' : n[i]-'a'+10;
sum = sum*radix+temp;
}
return sum;
}
long long find_radix(string n, long long num) {char it = *max_element(n.begin(), n.end());//查找最大元素的好办法long long low = (isdigit(it) ? it - '0': it - 'a' + 10) + 1;long long high = max(num, low);while (low <= high) {long long mid = (low + high) / 2;long long t = convert(n, mid); /// n->mid->t n->mid->tif (t < 0 || t > num) high = mid - 1;///1010->10->1010 1010->2->10 故t>num时说明当前进制大 应该往左找 t<0是溢出的情况 else if (t == num) return mid; else low = mid + 1;}return -1; } int main() {string n1, n2;long long tag = 0, radix = 0, result_radix;cin >> n1 >> n2 >> tag >> radix;result_radix = tag == 1 ? find_radix(n2, convert(n1, radix)) : find_radix(n1, convert(n2, radix));if (result_radix != -1) {printf("%lld", result_radix);} else {printf("Impossible");}return 0; }/** 没用二分17分 一堆答案错误 因为我没考虑好radix的范围 #include <iostream> #include <string> #include <fstream> #include <cstdio> using namespace std;int pow(int a, int n) {int ans = 1;for(int i=0; i<n; i++){ans *= a;}return ans; } int main() {int tag, radix;string a,b;int flag = 1;int aa[20], bb[20];long long int n1 = 0, n2 = 0;int ans = 0;cin >> a >> b >> tag >> radix;int cnt1 = 0, cnt2 = 0;for(int i=0; i<a.length(); i++){if(a[i]>='0' && a[i]<='9'){aa[cnt1++] = a[i]-'0';}else if(a[i]>='a' && a[i]<='z'){aa[cnt1++] = a[i] - '0' - 39;}}for(int i=0; i<b.length(); i++){if(b[i]>='0' && b[i]<='9'){bb[cnt2++] = b[i]-'0';}else if(b[i]>='a' && b[i]<='z'){bb[cnt2++] = b[i] - '0'-39;}} // for(int i=0; i<cnt2; i++){ // cout << bb[i] << endl; // }if(tag == 1){//to find 2nd's radix// convert 1st and 2nd to decimalfor(int i=0; i<cnt1; i++){n1 += aa[i]*pow(radix, cnt1-i-1);}for(ans = 2; ans <=36; ans++){n2 = 0;for(int i=0; i<cnt2; i++){n2 += bb[i]*pow(ans, cnt2-i-1);}if(n2 == n1){flag = 0;break;}}}else if(tag == 2){//to find 1st's radix//convert 1st and 2nd to decimalfor(int i=0; i<cnt2; i++){n1 += bb[i]*pow(radix, cnt2-i-1);}for(ans = 2; ans <=36; ans++){n2 = 0;for(int i=0; i<cnt1; i++){n2 += aa[i]*pow(ans, cnt1-i-1);}if(n2 == n1){flag = 0;break;}}}if(flag == 0)cout << ans;elsecout << "Impossible";return 0; } **/ /** 6 110 2 2 21 ab 1 2 **/
