Letters比赛第六场1001 Basically Speaking解题报告

1001 Basically Speaking(POJ 1546)

基本思路：基本的进制转换，先转化成十进制再转化成目标进制即可。

代码如下：

 

#include <cstdlib>
#include <iostream>
#include <cmath>

using namespace std;

int a = 0;
int b = 0;
char aNum[10];
char bNum[10];

void solve()
{
    int i = 7;
    int j = 0;
    int ans = 0;
    int alen = strlen(aNum);

    for(j=0; j<alen; j++)
    {
        if(aNum[j] >= 'A')
        {
            aNum[j] = aNum[j] - 'A' + 10;
        }
        else
        {
            aNum[j] = aNum[j] - '0';
        }
    }
    
    for(i=0; i<alen; i++)
    {
        ans = a * ans + aNum[i];
    }

    if(ans >= (int)pow((double)b, 7.0))
    {
        printf("  ERROR\n");
    }
    else
    {
        for(j=0; ans && j<7; j++)
        {
            bNum[j] = ans%b;
            ans = ans/b;
        }
        i = 7-j;
        while(i--)
        {
            printf(" ");
        }
        for(i=j-1; i>=0; i--)
        {
            printf("%0X", bNum[i]);
        }
        printf("\n");
    }
}

int main()
{
#ifdef MYLOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    while(scanf("%s %d %d", aNum, &a, &b) != EOF)
    {
        solve();  
    }

    return 0;
}