每日一题——第一百零九题

题目：进制转换合集。任意r进制与十进制之间的转换

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<ctype.h>
long stringToDecimal(const char* str, int base);
void convertToBaseR(int num, int r);int main() {char hexStr[] = "1A3F";char octStr[] = "123";char binStr[] = "101010";printf("十六进制转换成十进制为: %ld\n", stringToDecimal(hexStr, 16));printf("八进制转换成十进制为: %ld\n", stringToDecimal(octStr, 8));printf("二进制转换成十进制为: %ld\n", stringToDecimal(binStr, 2));int num, r;printf("请输入十进制的整数: ");scanf_s("%d", &num);printf("请输入要转化的进制数: ");scanf_s("%d", &r);convertToBaseR(num, r);return 0;
}/// <summary>
/// 将字符串str转为十进制
/// </summary>
/// <param name="str">字符串</param>
/// <param name="base">基数2， 8， 16等</param>
/// <returns></returns>
long stringToDecimal(const char* str, int base) {long result = 0;int multiplier = 1;int length = strlen(str);//从字符串末尾开始解析for (int  i = length - 1; i >= 0; i--){char ch = tolower(str[i]);int digit = -1;//根据字符计算其对应的值if (ch >= '0' && ch <= '9') {digit = ch - '0';//将字符串转换成对应的数字}else if(base == 16 && (ch >= 'a' && ch <= 'f')){digit = ch - 'a' + 10;}else if(ch < '0' || (ch > '9' && base !=16) ||(base == 16) && ch > 'f'){//出错return 0;}//判断是否在有效范围内if (digit >= base) {return 0;}result += digit * multiplier;  //更新累加结果multiplier *= base;//更新下一次计算的基数}}/// <summary>
/// 将十进制数转换成任意的r进制数
/// </summary>
/// <param name="num">数字</param>
/// <param name="r">要转换的目标进制数</param>
void convertToBaseR(int num, int r) {int index = 0;char result[65];//假设转换后的数不超过64位//处理num = 0的情况if (num == 0) {printf("0\n");return;}while (num > 0){int remainder = num % r;if (remainder < 10) {result[index++] = '0' + remainder;//余数小于10的,将其转换成0~9之间的字符}else{result[index++] = 'A' + (remainder - 10);//余数大于10的，将其转换成'A' ~ 'Z' 字符}num = num / r; //更新下一次参与计算的数字}result[index] = '\0'; //在字符串末尾添加结束符//因为结果是从低位到高位存储的，所以需要反向打印for (int i = index; i >= 0; i--){printf("%c", result[i]);}printf("\n");
}