HDU 又一版 A+B

来源：互联网发布：下载itunes软件编辑：程序博客网时间：2024/05/01 14:46

题目：输入两个不超过整型定义的非负10进制整数A和B(<=2³¹-1)，输出A+B的m (1 < m <10)进制数。

http://acm.hdu.edu.cn/showproblem.php?pid=1877

// LANG: C++#include <stdio.h>#include <string.h>#include <stdlib.h>//http://acm.hdu.edu.cn/showproblem.php?pid=1877//输入两个不超过整型定义的非负10进制整数A和B(<=231-1)，输出A+B的m (1 < m <10)进制数。//进制转换，base进制数void work(unsigned int num, int base){char str[40] = {0};int i = 0;// 判断结果是否为0if (!num){printf("0\n");return ;}while (num){str[i++] = num % base + '0';num /= base;}--i;while (i >= 0) putchar(str[i--]);putchar('\n');}//多进制转换，num=源数据，base进制数,buffer输出数据void work2(unsigned int num, int base,char* buffer){char str[40] = {0};int i = 0;int j=0;// 判断结果是否为0if (!num){printf("0\n");return ;}while (num){str[i++] = num % base + '0';num /= base;}--i;while (i >= 0){buffer[j++]=str[i--];}//printf("%s\n",buffer);}// 2 1 7int main(int argc, char **argv){unsigned int base, left, right;while (1){scanf ("%u", &base);if (!base) break;scanf("%u %u", &left, &right);left += right;char* buffer;buffer=(char*)malloc(sizeof(char*)*1024);memset(buffer,0,sizeof(char*)*1024);work2(left, base,buffer);printf("%s\n",buffer);}return 0;}

0 0