FZU 2102 Solve equation 多进制计算

Solve equation

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

FZU 2102

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)

每一位的数要乘进制 不要只是乘了10

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INF     0x3f3f3f3f
#define INFLL   0x3f3f3f3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int,int> PII;

const int MAXN = 1000 + 5;

char A[MAXN];
char B[MAXN];

int getnum(char a){
   if(a >= '0' && a <= '9')  return a - '0';
   else return a - 'a' + 10;
}

int main()
{
    //FIN
    int T, C;
    scanf("%d", &T);
    while(T--){
        scanf("%s%s", A, B);
        scanf("%d", &C);
        int lenA = strlen(A);
        int lenB = strlen(B);
        int numA = 0, numB = 0;
        for(int i = 0;i < lenA; i ++){
            numA = numA * C + getnum(A[i]);
        }
        for(int i = 0;i < lenB; i ++){
            numB = numB * C + getnum(B[i]);
        }
        //cout<<numA<<"       "<<numB<<endl;
        int d = numA % numB;
        int k = numA / numB;
        printf("(%d,%d)
", k, d);
    }

}