题解 | 进制转换

//本题位数较大,超出了long long范围,而在十进制转二进制的过程中需不断除进制2,这个过程需借助字符串处理
//从最高位开始处理,currnet表示当前正在处理的位, (current / 进制) 得出商对应的当前位,若有余数,则余数乘10并加上下一位,循环遍历完所有的字符,remain初始化为0
#include <iostream>
#include <vector>
#include <string>
using namespace std;

string divide(string str) {
    int remain = 0;
    for (int i = 0; i < str.size(); i++) {
        int current = remain * 10 + str[i] - '0';
        str[i] = current / 2 + '0';
        remain = current % 2;
    }
    int pos = 0;
    while (str[pos] == '0') {
        pos++;
    }
    return str.substr(pos);
}
int main() {
    string str;
    // while (scanf("%s", &str) != EOF) {
    while (cin >> str) {
        // if (str == "0") {
        //     printf("0");
        // }
        vector <int> vec;
        while (str!="") {      //一般是除数为0时跳出循环,但由于除法去掉了前导0,就会返回空串
            int remain = str[str.size() - 1]-'0';
            vec.push_back(remain % 2);
            str = divide(str);
        }

        for (int i = vec.size() - 1; i >= 0; i--) {
            printf("%d", vec[i]);
            //vec.pop_back();
        }

        printf("\n");
    }
    return 0;
}