poj 1220 NUMBER BASE CONVERSION(短除法进制转换）

题目连接：1220 NUMBER BASE CONVERSION

题目大意：给出两个进制oldBase 和newBase, 以及以oldBase进制存在的数。要求将这个oldBase进制的数转换成newBase进制的数。

解题思路：短除法，只不过时直接利用了高精度的除法运算， 并且将以前默认的*10换成的*oldBase。

#include <stdio.h>
#include <string.h>
const int N = 1005;
int newBase, oldBase, n, cnt, num[N];
char str[N];

int getnum(char c) {
    if (c >= '0' && c <= '9')
	return c - '0';
    else if (c >= 'A' && c <= 'Z')
	return c - 'A' + 10;
    else
	return c - 'a' + 36;
}

char getchar(int c) {
    if (c >= 0 && c <= 9)
	return '0' + c;
    else if (c >= 10 && c < 36)
	return 'A' + c - 10;
    else
	return 'a' + c - 36;
}

void change() {
    memset(num, 0, sizeof(num));
    n = strlen(str);
    for (int i = 0; i < n; i++)
	num[i] = getnum(str[i]);
}

void solve() {
    int flag = 1;
    memset(str, 0, sizeof(str));
    cnt = 0;

    while (flag) {
	int t = 0, k;
	flag = 0;
	for (int i = 0; i < n; i++) {
	    k = num[i] + t * oldBase;
	    num[i] = k / newBase;
	    if (num[i])	flag = 1;
	    t = k % newBase;
	}
	str[cnt++] = getchar(t);
    }
}

int main() {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
	scanf("%d%d%s", &oldBase, &newBase, str);

	printf("%d %s
", oldBase, str);

	change();
	solve();

	printf("%d ", newBase);
	for (int i = cnt - 1; i >= 0; i--)
	    printf("%c", str[i]);
	printf("
");
	if (cas)    printf("
");
    }
    return 0;
}