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;}