/*
思路
    思路一：
        1. string先接收再char数组接收
        2. 每一位的数字乘以它的（数位减一次方）
*/
/*改进
#include <bits/stdc++.h>
using namespace std;

int main()
{
    string str;
    int n;
    cin >> str >> n;
    vector<char> ch;
    for ( char c : str ) {
        ch.push_back(c);
    }
    vector<char> chs(ch.size());
    int j = 0;
    for ( int i = ch.size()-1; i >= 0; i-- ) {
        chs[j] = ch[i];
        j++;
    }
    int count = 0;
    for ( size_t i = 0; i < chs.size(); i++ ) {
        count += (chs[i] - 48) * pow(n,i);
        
    }
    cout << count << endl;


    return 0;
}*/
#include <bits/stdc++.h>
using namespace std;

int main()
{
    string numStr;
    int base;
    cin >> numStr >> base;

    //创建动态数组 digits(数位)
    vector<char> digits(numStr.size());
    for ( size_t i = 0; i < digits.size(); i++ ) {
        digits[i] = numStr[numStr.size()-1-i];
    }

    //计算进制
    int result = 0;
    int power = 1;
    for ( size_t i = 0; i < digits.size(); i++ ) {
        if ( digits[i] >= 'A' && digits[i] <= 'F' ) {
            result += (digits[i] - 'A' + 10) * power;
        }else {
            result += (digits[i] - '0') * power;
        }
        power *= base;
    }

    cout << result << endl;

    return 0;
}

#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef pair<int,int> aII;  
using ll = long long;
using ULL = unsigned long long;
const int N = 1e6+5;

ll m;
string n;
inline void solve() { 
    cin >> n >> m;
    ll res = 0;
    for (auto &i: n) {
        res = res * m + (isdigit(i)? i - '0' : i - 'A' + 10);
    }
    cout << res << endl;
}
int main() { 
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int _ = 1; 
    //int _; cin >> _;
    while (_--) solve();
    return 0;
}

// https://dashoj.com/p/96
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

string s;
int base, power = 0;
ll result = 0;

int main() {
	cin >> s >> base;
	for (int i = s.size() - 1; i >= 0; i--) {
		int value;
		if (isdigit(s[i])) value = s[i] - '0';
		else value = s[i] - 'A' + 10;
		result += value * pow(base, power);
		power++;
	}
	cout << result << endl;
	return 0;
}

用 1001 分别以二进制、八进制、十六进制转成十进制

• 二进制 1001 转十进制

  • $1 * 2^{0}+0 * 2^{1}+0 * 2^{2}+1 * 2^{3}$

• 八进制 1001 转十进制

  • $1 * 8^{0}+0 * 8^{1}+0 * 8^{2}+1 * 8^{3}$

• 十六进制 1001 转十进制

  • $1 * 16^{0}+0 * 16^{1}+0 * 16^{2}+1 * 16^{3}$

对于小数, 则是 $a * K^{-1} + a * K^{-2} + ...$

输入一个 $x$ 进制数, 把它转化为 十进制!

#include <iostream>
#include <string>

using namespace std;

int main() {
	string str;
	int jz;
	cin >> str >> jz;
	
	long long res = 0;
	long long k = 1;
	for (int i = str.size() - 1; i >= 0; --i) {
		if ('0' <= str[i] && str[i] <= '9') {
			res += (str[i] - '0') * k;
		} else {
			res += (10 + str[i] - 'A') * k;
		}
		k *= jz;
	}
	
	cout << res << endl;
	
	return 0;
}

#include <bits/stdc++.h>

using namespace std; int main(){ string s; long long ans=0;//必须让自定义函数等于0 否则就会再下列计算中 自己加上自己的asall值 int k=0, base=0; cin>> s >> base;

for ( int i=s.size()-1;i>=0; i--){ //缺失了条件 if (s[i]>='A' ){ ans += ( s[i]-'A'+10 )* pow( base, k++); } else { ans += (s[i]-'0')* pow( base,k++); }

} cout << ans<<endl;

return 0;

}