进制转换（将M进制的数X转换为N进制的数输出）
题目
#include <stdio.h>#include <stdlib.h>#include <string.h> #define LEN 2000 char str[LEN], another[LEN];int ten[LEN]; int switchToTen();void switchToAnother(int k, int n); int main(){ int m, n, k; while (scanf("%d %d", &m, &n) != EOF) { scanf("%s", str); k = switchToTen(m); switchToAnother(k, n); } return 0;} int switchToTen(int m){ int i, j, len, k, c; //初始化 len = strlen(str); k = 1; memset(ten, 0, sizeof(ten)); //转换为10进制数 for (i = 0; i < len; i ++) { for (j = 0; j < k; j ++) { ten[j] *= m; } if (str[i] >= '0' && str[i] <= '9') { ten[0] += str[i] - '0'; }else if (str[i] >= 'A' && str[i] <= 'Z') { ten[0] += str[i] - 'A' + 10; }else if (str[i] >= 'a' && str[i] <= 'z') { ten[0] += str[i] - 'a' + 10; } for (j = c = 0; j < k; j ++) { ten[j] += c; if (ten[j] >= 10) { c = ten[j] / 10; ten[j] %= 10; }else { c = 0; } } while (c) { ten[k ++] = c % 10; c /= 10; } } //翻转数组 int temp; for (i = 0, j = k - 1; i < j; i ++, j --) { temp = ten[i]; ten[i] = ten[j]; ten[j] = temp; } return k;} void switchToAnother(int k, int n){ int sum, i, r, t, d; sum = 1; r = 0; memset(another, 0, sizeof(another)); while (sum) { sum = 0; for (i = 0; i < k; i ++) { d = ten[i] / n; sum += d; if (i == k - 1) { t = ten[i] % n; if (t >= 0 && t <= 9) { another[r] = t + '0'; }else { another[r] = t - 10 + 'a'; } r ++; }else { ten[i + 1] += ten[i] % n * 10; } ten[i] = d; } } //打印是输出 for (i = r - 1; i >= 0; i --) { printf("%c", another[i]); } printf("\n");}/************************************************************** Problem: 1080 User: wangzhengyi Language: C Result: Accepted Time:170 ms Memory:920 kb****************************************************************/