题解 | 进制转换

#include <cmath>
#include <iostream>
using namespace std;
#define I int
#define LL long long

void jzzh() ;

int main() {
	jzzh();
	
	return 0;
}

void jzzh() { //进制转换
	I n = 0, ni = 0, res = 0;
	char c[12];
	while (cin >> c[n]) {		//输入n+1个字符
		n++;
	}
	ni = n - 2;					//n-1位数字
	for (I i = 2; i < n; i++) {	//从ni=2开始
		if (c[i] <= '9')
			res += (c[i] - '0') * pow(16, --ni);
		else if (c[i] >= 'A')
			res += (c[i] - '7') * pow(16, --ni);
	}
	cout << res;
}