PAT 1010 Radix

过了一天感觉没干什么，想刷一发题弥补一下，正在考虑去做哪道，室友说去试试PAT 1010，果然当年自己直接跳过了这题，一看这通过率（0.07）有点夸张了。题目是已知一个数，求另外一个数的基数，使得这两个数数值上相等。很自然的考虑到使用二分搜索来确定这个基数，数字表示使用[0-9a-z],这tmd的让人很容易的想到基数的范围就在1~36之间了，艹，基数是可以超过这个范围的，如果没有考虑到这一点，可以得到的一个典型分值就是19分。不过基数在[1, 36]之间的话这个搜索范围太小了，直接暴力遍历也可以，于是闷声不响的扩大范围，总之是坑题，哎没办法，老女人就是爱这样。下面给出代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;

char char2num[128];

int remove_leading_zero(char* str) {
    if (str[0] == ‘\0‘) return 0;
    int ri = 0, wi = 0;
    while (str[ri] == ‘0‘ || str[ri] == ‘+‘ || str[ri] == ‘-‘) ri++;
    int len = 0;
    while ((str[wi++] = str[ri++]) != ‘\0‘) len++;
    if (len == 0) {
        str[0] == ‘0‘;
        str[++len] = ‘\0‘;
    }
    return len;
}

long long value(const char* str, int len, long long radix) {
    long long ret = 0;
    long long r = 1;
    for (int i=len - 1; i>=0; i--) {
        int digit = char2num[str[i]];
        // we should check the number validation
        if (digit >= radix) return -1;
        ret += r * digit;
        r *= radix;
    }
    return ret;
}

int inc_cmp(char* str, int len, long long radix, long long target){
    long long v = 0;
    long long r = 1;

    for (int i=len - 1; i>=0; i--) {
        int digit = char2num[str[i]];
        v += r * digit;
        r *=radix;
        if (v > target) {
            return 1;
        }
    }
    if (v == target) {
        return 0;
    } else {
        return -1;
    }
}

long long binary_search(char* str, int len, long long lo, long long hi, long long target){
    long long mid;
    lo = lo - 1;
    hi = hi + 1;

    while(lo + 1< hi){
        mid = (lo + hi) / 2;
        int res = inc_cmp(str, len, mid, target);
        if(res > 0) {
            hi = mid;
        } else if(res < 0) {
            lo = mid;
        } else {
            return mid;
        }
    }
    return -1;
}

int main() {
    // init char2num lookup table
    for (int i=0; i<10; i++) char2num[‘0‘ + i] = i;
    for (int i=‘a‘; i<=‘z‘; i++) char2num[i] = i - ‘a‘ + 10;

    char num1[16] = {‘\0‘};
    char num2[16] = {‘\0‘};
    char *pnum1 = num1, *pnum2 = num2;
    int tag = 0;
    long long bradix = 0;

    scanf("%s%s%d%ld", num1, num2, &tag, &bradix);

    // we always assure that bradix is the radix of pnum1
    // and pnum2 is which we should guess its radix
    if (tag != 1) {
        pnum1 = num2;
        pnum2 = num1;
    }

    int n1len = remove_leading_zero(pnum1);
    int n2len = remove_leading_zero(pnum2);

    long long n1 = value(pnum1, n1len, bradix);

    bool is_same = !strcmp(pnum1, pnum2);

    if(is_same) {
        if (n1len > 1) {
            // must be same radix, if digits more than one
            printf("%d\n", bradix);
        } else {
            // only one digit, so choose a smallest valid radix
            printf("%d\n", n1 + 1);
        }
        return 0;
    } 

    long long lo = 0;
    for (int i=0; i<n2len; i++) {
        int d = char2num[pnum2[i]];
        if (d + 1> lo) lo = d + 1;
    }

    long long hi = n1 > lo ? n1 : lo;
    int res = binary_search(pnum2, n2len, lo, hi, n1);
    if (res < 0) {
        printf("Impossible\n");
    } else {
        printf("%d", res);
    }

    return 0;
}