FZU Problem 2102 Solve equation题解

Problem 2102 Solve equation

Accept: 883    Submit: 2075
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

32bc 33f 16123 100 101 1 2

Sample Output

(0,700)(1,23)(1,0)

Source

“高教社杯”第三届福建省大学生程序设计竞赛

一道水题~~~就是考查进制转换，，比赛时总是想着用16进制输入输出，陷入了误区，，，输入各种进制，应用字符串，然后再转化为十进制数~~~~

代码详解：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
int judge(char p[],int x)
{
    int len=strlen(p),ans=0,k,l=1;
    for(int i=len-1;i>=0;i--)
    {
        if(p[i]>='0'&&p[i]<='9')
            k=p[i]-'0';
        else
            k=p[i]-'a'+10;
        ans+=l*k;
        l*=x;
    }
    return ans;
}
int main()
{
    int t,c;
    scanf("%d",&t);
    while(t--)
    {
        int ans1,ans2,ans3,ans4;
        char s1[105],s2[105];
        scanf("%s%s%d",s1,s2,&c);
        ans1=judge(s1,c);//调用函数转换
        ans2=judge(s2,c);
        ans3=ans1/ans2;
        ans4=ans1%ans2;
        printf("(%d,%d)\n",ans3,ans4);
    }
    return 0;
}