基础练习

十六进制转八进制1：

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

long long pow(long long x, long long y) {
	long long ans = 1;
	while (y > 0) {
		if (y & 1) {
			ans *= x;
		}
		x *= x;
		y >>= 1;
	}
	return ans;
}

int main() {
	int t;
	cin >> t;
	string s1, s2;
	long long n;
	while (t--) {
		cin >> s1;
		s2 = "";
		n = 0;
		for (int i = 0; i < s1.size(); i++) {
			if (s1[i] <= '9' && s1[i] >= '0') {
				n += (s1[i] - '0') * pow(16, s1.size() - i - 1);
			} else {
				n += (s1[i] - 'A' + 10) * pow(16, s1.size() - i - 1);
			}
		}
		int a, b;
		while (n) {
			a = n % 8;
			s2 += (a + '0');
			n /= 8;
		}
		for (int i = s2.size() - 1; i >= 0; i--) {
			cout << s2[i];
		}
		cout << endl;
	}
	return 0;
}

十六进制转八进制2：

#include<iostream>
using namespace std;

//思路：100000位的十六进制数，这么大的数不好直接处理，以二进制字符串转换为八进制即可 
int main(){
	int n;
	cin>>n;
	while(n--){
		string s;
		cin>>s;
		
		int len1 = s.length();
		string res1="";
		//将16进制转换为二进制字符串 
		for (int i=0;i<len1;i++){
			switch(s[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;
			}
		}
		//二进制字符串每四个数就是一个16进制数，每3个数就是一个8进制数，这里可能存在缺0的情况，需要补充
		int len2=res1.length() ;
		switch(len2%3){
			case 1: res1 = "00"+res1; break;
			case 2: res1 = "0"+res1; break;
		}
		
		//此时的res1就是一个合法的二进制表示8进制的字符串了,下面开始生成八进制字符串了 
		string res2="";
		len2=res1.length();
		for (int i=0;i<len2;i+=3){
			string t = res1.substr(i,3); 
			// 避免因为000带来导致转为为八进制时候有前导0 
			if (i==0&&t=="000") res2+="";
			else res2 += (4*((t[0])-'0')+ 2*((t[1])-'0' )+ ((t[2])-'0'))+'0';
			
		}
		cout<<res2<<endl;
	} 
	
	return 0;
}

特殊回文数：枚举暴力+小优化

#include <iostream>
using namespace std;

//枚举暴力+小优化
int main() {
	int n;
	cin >> n;
	//枚举五位数
	for (int i = 1; i <= 9; i++) {
		for (int j = 0; j <= 9; j++) {
			for (int k = 0; k <= 9; k++) {
				if (i * 2 + j * 2 + k == n) {
					cout << i * 10000 + j * 1000 + k * 100 + j * 10 + i << endl;
				}
			}
		}
	}
	//枚举六位数
	for (int i = 1; i <= 9; i++) {
		for (int j = 0; j <= 9; j++) {
			for (int k = 0; k <= 9; k++) {
				if (i * 2 + j * 2 + k * 2 == n) {
					cout << i * 100000 + j * 10000 + k * 1000 + k * 100 + j * 10 + i << endl;
				}
			}
		}
	}
	return 0;
}