FZOJ 2102 Solve equation

                                                                                                                                     Problem 2102 Solve equation

Accept: 1097    Submit: 2608
Time Limit: 1000 mSec    Memory Limit :
32768 KB

Problem Description

You are given two positive integers A and B in Base C. For the equation:

A=k*B+d

We know there always existing many non-negative pairs (k, d) that satisfy the
equation above. Now in this problem, we want to maximize k.

For example, A="123" and B="100", C=10. So both A and B are in Base 10. Then
we have:

(1) A=0*B+123

(2) A=1*B+23

As we want to maximize k, we finally get one solution: (1, 23)

The range of C is between 2 and 16, and we use 'a', 'b', 'c', 'd', 'e', 'f'
to represent 10, 11, 12, 13, 14, 15, respectively.

Input

The first line of the input contains an integer T (T≤10), indicating the
number of test cases.

Then T cases, for any case, only 3 positive integers A, B and C (2≤C≤16) in a
single line. You can assume that in Base 10, both A and B is less than
2^31.

Output

For each test case, output the solution “(k,d)” to the
equation in Base 10.

Sample Input

3
2bc 33f 16
123 100 10
1 1 2

Sample Output

(0,700)
(1,23)
(1,0) 

 

题意：对于形式A=k*B+d，给定A，B两个数，求k,d两个值，且A，B的进制有可能是2~16进制中的其中一种

思路：首先将A，B都换算成10进制，再作除法运算。

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;

int base;

int translator(string s) {
    int sum = ;
    for (int i = ; i < s.size();i++) {
        if (s[i] >= ''&&s[i] <= '')
            sum = sum*base + s[i] - '';
        else
            sum = sum*base + s[i] - 'a' + ;
  }
    return sum;
}

int main() {
    int T;
    scanf("%d",&T);
    while (T--) {
        string s1, s2;
        cin >> s1 >> s2;
        scanf("%d",&base);
        int a = translator(s1);
        int b = translator(s2);
        int c = a / b;
        int mod = a%b;
        printf("(%d,%d)\n",c,mod);

    }
    return ;
}