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1010

时间:2019-05-11 13:34:32      阅读:158      评论:0      收藏:0      [点我收藏+]

标签:tag   number   ati   rbegin   sum   let   dex   har   lse   

#include <iostream>
#include <cctype>
#include <algorithm>
#include <cmath>
using namespace std;
long long convert(string n, long long radix) {
    long long sum = 0;
    int index = 0, temp = 0;
    for (auto it = n.rbegin(); it != n.rend(); it++) {
        temp = isdigit(*it) ? *it - ‘0‘ : *it - ‘a‘ + 10;// - ‘a‘ + 10 transform 16radix letter to 10radix number;-‘0‘transform string to number;
        sum += temp * pow(radix, index++);
    }
    return sum;
}
long long find_radix(string n, long long num) {
    char it = *max_element(n.begin(), n.end());
    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);
        if (t < 0 || t > num) //if t is negative number,that mean t out of long long boundary.
            high = mid - 1;
        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;
}

1010

标签:tag   number   ati   rbegin   sum   let   dex   har   lse   

原文地址:https://www.cnblogs.com/masayoshi/p/10848297.html

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