POJ1220（大数进制转换）

/*
ID: LinKArftc
PROG: 1220.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

const int maxn = 1000;
char str[maxn];//输入字符串
int start[maxn], ans[maxn], res[maxn];//被除数，商，余数
int oldBase, newBase;//转换前后进制

//单个字符得到数字
int getNum(char c) {//这里的进制字符是先数字，后大写字母，后小写字母
    if (c >= ‘0‘ && c <= ‘9‘) return c - ‘0‘;//数字
    if (c >= ‘A‘ && c <= ‘Z‘) return c - ‘A‘ + 10;//大写字母
    return c - ‘a‘ + 36;
}
//数字得到字符
char getChar(int i) {
    if (i >= 0 && i <= 9) return i + ‘0‘;
    if (i >= 10 && i <= 35) return i - 10 + ‘A‘;
    return i - 36 + ‘a‘;
}
//把输入的字符串的各个数位还原成数字形式
void change() {
    start[0] = strlen(str);//数组的0位存的是数组长度
    for (int i = 1; i <= start[0]; i ++) start[i] = getNum(str[i-1]);
}

void solve() {
    memset(res, 0, sizeof(res));//余数位初始化为空
    int y, i, j;
    while (start[0] >= 1) {
        y = 0; i = 1;
        ans[0] = start[0];
        while (i <= start[0]) {
            y = y * oldBase + start[i];
            ans[i ++] = y / newBase;
            y %= newBase;
        }
        res[++ res[0]] = y;//这一轮得到的余数
        i = 1;//找下一轮商的起始处，去掉前面的0
        while ((i <= ans[0]) && (ans[i] == 0)) i ++;
        memset(start, 0, sizeof(start));
        for (j = i; j <= ans[0]; j ++) start[++ start[0]] = ans[j];
        memset(ans, 0, sizeof(ans));
    }
}

void output() {//从高位到低位逆序输出
    for (int i = res[0]; i >= 1; i --) printf("%c", getChar(res[i]));
    printf("\n");
}

int main() {
    //input;
    int T;
    scanf("%d", &T);
    while (T --) {
        scanf("%d %d %s", &oldBase, &newBase, str);
        printf("%d %s\n", oldBase, str);
        change();
        solve();
        printf("%d ", newBase);
        output();
        printf("\n");
    }
    return 0;
}