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;    elsereturn 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;    elsereturn '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\n", oldBase, str);change();solve();printf("%d ", newBase);for (int i = cnt - 1; i >= 0; i--)    printf("%c", str[i]);printf("\n");if (cas)    printf("\n");    }    return 0;}