PAT甲级1010

1010. Radix (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
LL Map[256];//0~9、a~z与0~35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
}
r
4000
eturn 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;
}