PAT 甲级 1010. Radix

原题传送门

• 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