# UVa 355
### 題目連結：[UVa355](http://domen111.github.io/UVa-Easy-Viewer/?355)
### [參考1](http://chchwy.blogspot.com/2010/01/acmuva-355-bases-are-loaded.html)
:::spoiler sample code

#include<iostream>
#include<algorithm>
using namespace std;

inline int CharToDigit(char c)
{
    if ( isdigit(c) )
        return c - '0';
    return c - 'A' + 10;
}

inline char DigitToChar(int n)
{
    return "0123456789ABCDEF"[n];
}

bool checkIllegal(int base1, string& in )
{
    for (int i = 0; i < in.length(); ++i)
        if ( CharToDigit(in[i]) >= base1 )
            return true;
    return false;
}

/* convert string 'in' to int 'value'. */
long long parseValue( int base1, string& in )
{
    if ( checkIllegal(base1, in) ) return -1;

    long long value = 0;
    long long curBase = 1;

    for (int i = in.length() - 1; i >= 0; --i)
    {

        value += CharToDigit(in[i]) * curBase;
        curBase *= base1;
    }
    return value;
}

/* convert int 'value' to string 'out' */
string toBase( int base2, long long value )
{
    if (value == 0) return "0";

    string out;
    while ( value > 0 )
    {
        out += DigitToChar( value % base2 );
        value /= base2;
    }
    reverse(out.begin(), out.end());
    return out;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("355.in", "r", stdin);
#endif

    int base1, base2;
    string in;

    while ( cin >> base1 >> base2 >> in )
    {
        long long value = parseValue(base1, in);

        if ( value < 0 ) //wrong
        {
            cout << in << " is an illegal base " << base1 << " number\n";
            continue;
        }
        cout << in << " base " << base1 << " = " << toBase(base2, value) << " base " << base2 << "\n";
    }
    return 0;
}

:::
### [參考2](http://onlinejudgesolution.blogspot.com/2017/05/uva-solution-355-bases-are-loaded.html)
:::spoiler sample code

#include <stdio.h>

int main() {
    int a, b;
    char s[20];
    while(scanf("%d %d %s", &a, &b, s) == 3) {
        int i, flag = 0;
        long long dec = 0, base = 1;
        for(i = 0; s[i]; i++) {
            dec *= a;
            if(s[i] >= '0' && s[i] <= '9') {
                if(s[i]-'0' >= a) {
                    flag = 1;
                } else {
                    dec += s[i]-'0';
                }
            } else {
                if(s[i]-'A'+10 >= a) {
                    flag = 1;
                } else {
                    dec += (s[i]-'A'+10);
                }
            }
        }
        if(flag) {
            printf("%s is an illegal base %d number\n", s, a);
        } else {
            long long c[50] = {}, ct = -1;
            while(dec) {
                c[++ct] = dec%b;
                dec /= b;
            }
            if(ct < 0) ct = 0;
            printf("%s base %d = ", s, a);
            while(ct >= 0) {
                printf("%c", c[ct] <= 9 ? c[ct]+'0' : c[ct]-10+'A');
                ct--;
            }
            printf(" base %d\n", b);
        }
    }
    return 0;
}

:::
### 題述：
#### 寫一個程式做進位制之間的轉換（2進位到16進位）。其中A代表10，B代表11......，F代表15。
#### 每組測試資料一列，有3個值。第一個值為一個正整數m，代表要轉換的這個數是幾進位的數。第二個值為一個正整數n，代表要把這個數轉換成幾進位的數。第三個值就是要轉換的數（m進位），這個值最長不會超過10個字元的長度，且有可能在m進位之下是不正確的（例如Sample Input中的第二列，126不是一個正確的5進位數）。
#### 以Sample Input的第一列為例說明：要把2進位表示法的10101轉換成10進位的表示法。
### c++ code：

:::success
**``sample input``**
2 10 10101
5 3 126
15 11 A4C
5 15 0
:::
:::success
**``sample output``**
10101 base 2 = 21 base 10
126 is an illegal base 5 number
A4C base 15 = 1821 base 11
0 base 5 = 0 base 15
:::
#### [返回首頁](https://hackmd.io/@fkleofk/APCS#355)
###### tags: `APCS選修` `C++` `UVa`