进制转换

快捷方法

printf("%o", x); // 以8进制输出
printf("%d", x); // 以10进制输出
printf("%x", x); // 以16进制输出

cin >> hex >> x; // hex以16进制形式读入
cout << x; // 默认以10进制形式输出

cout << hex << x; // 以16进制输出
cout << dec << x; // 以10进制输出
cout << oct << x; // 以8进制输出
cout << bitset<8>(x); // 以2进制输出保留<8>位输出

库函数

itoa(num, str, 2); // num是一个int型的，是要转化的10进制数，str是转化结果，后面的值为目标进制。

#include<cstdio> 
#include<cstdlib> 
int main()  
{  
    int num = 10;  
    char str[100];  
    _itoa(num, str, 2);
    printf("%s\n", str);  
    return 0;  
}

来源连接 https://blog.csdn.net/vir_lee/article/details/80645066

16进制转8进制

#include <iostream>
#include <cstring>

using namespace std;

string ans[15];

int main()
{
    int n;
    cin >> n;

    while (n --)
    {
        string str;
        cin >> str;

        string res1 = "";
        // 转为二进制 
        for (int i = 0; i < str.size(); i ++ )
        {
            switch(str[i])
            {
                case '0' : res1 += "0000"; break;
                case '1' : res1 += "0001"; break;
                case '2' : res1 += "0010"; break;
                case '3' : res1 += "0011"; break;
                case '4' : res1 += "0100"; break;
                case '5' : res1 += "0101"; break;
                case '6' : res1 += "0110"; break;
                case '7' : res1 += "0111"; break;
                case '8' : res1 += "1000"; break;
                case '9' : res1 += "1001"; break;
                case 'A' : res1 += "1010"; break;
                case 'B' : res1 += "1011"; break;
                case 'C' : res1 += "1100"; break;
                case 'D' : res1 += "1101"; break;
                case 'E' : res1 += "1110"; break;
                case 'F' : res1 += "1111"; break;
            }
        }

        switch (res1.size()%3)
        {
            case 1: res1 = "00" + res1; break;
            case 2: res1 = "0" + res1; break;
        }

        string res2 = "";
        for (int i = 0; i < res1.size(); i += 3)
        {
            string t = res1.substr(i, 3);

            if (i == 0 && t == "000") continue;
            else res2 += (4*(t[0] - '0') + 2 * (t[1] - '0') + (t[2] - '0')) + '0';
        }
        cout << res2 << endl;
    }

    return 0;
}

10进制转16进制

#include <iostream>

using namespace std;

typedef long long LL;

char a[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};

int main()
{
    LL n;
    cin >> n;

    string str = "";
    if (n == 0) puts("0");
    else
    {
        while (n)
        {
            str += a[n%16];
            n /= 16;
        }
    }

    for (int i = str.size() - 1; i >= 0; i -- ) cout << str[i];

    return 0;
}