- 目前来讲最难的,用到了二分思想
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
LL Map[256]; // 0~9、a~z与10~35的对应
LL inf = (1LL << 63) - 1; // long long的最大值2^63-1,注意加括号
void init() {
for (char c = '0'; c <= '9'; c++) {
Map[c] = c - '0';
}
for (char c = 'a'; c <= 'z'; c++) {
Map[c] = c - 'a' + 10; // 将'a'~'z'映射到10~35
}
}
LL convertNum10(char a[], LL radix, LL t) { // 将a转换为十进制,t为上界
LL ans = 0;
int len = strlen(a);
for (int i = 0; i < len; i++) {
ans = ans*radix + Map[a[i]]; // 进制转换
if (ans<0 || ans>t) return -1; // 溢出或超过N1的十进制
}
return ans;
}
int cmp(char N2[], LL radix, LL t) { // N2的十进制与t进行比较
int len = strlen(N2);
LL num = convertNum10(N2, radix, t); // 将N2转换为十进制
if (num < 0) return 1; // 溢出肯定是N2>t;
if (t > num)return -1; // t较大,返回-1
else if (t == num)return 0; // 相等,返回0
else return 1; // num较大,返回1
}
LL binarySearch(char N2[], LL left, LL right, LL t) { // 二分求解N2的进制
LL mid;
while (left <= right) {
mid = (left + right) / 2;
int flag = cmp(N2, mid, t); // 判断N2转换为十进制后与t比较
if (flag == 0)return mid; // 找到解,返回mid
else if (flag == -1)left = mid + 1;// 往右子区间继续查找
else right = mid - 1; // 往左子区间继续查找
}
return -1;
}
int findLargestDigit(char N2[]) { // 求最大的数位
int ans = -1, len = strlen(N2);
for (int i = 0; i < len; i++) {
if (Map[N2[i]] > ans) {
ans = Map[N2[i]];
}
}
return ans + 1; // 最大的数位为ans,说明进制数的底线是ans+1
}
char N1[20], N2[20], temp[20];
int tag, radix;
int main() {
init();
scanf("%s %s %d %d", N1, N2, &tag, &radix);
if (tag == 2) {
strcpy(temp, N1);
strcpy(N1, N2);
strcpy(N2, temp);
}
LL t = convertNum10(N1, radix, inf); // 将N1从radix进制转换为十进制
LL low = findLargestDigit(N2); // 找到N2中数位最大的位加1,当成二分下界
LL high = max(low, t) + 1; // 上界
LL ans = binarySearch(N2, low, high, t); // 二分
if (ans == -1)
printf("Impossible\n");
else
printf("%lld\n", ans);
return 0;
}
附原题:
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