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英文字母ASCII相互轉換

(1)  (2)  (3)  (TA)

解法1(上傳Judge)

#include<iostream>
using namespace std;
int main() {
	char s[2];
	cin >> s;
	if(s[1] == 0)
		cout << (int)s[0] << endl;
	else
		cout << (char)((int)s[0]*10+(int)s[1]-16) << endl;
	return 0;
}

解法2

#include<iostream>
using namespace std;
int main() {
	char s[2];
	cin >> s;
	if(s[1] == 0)
		cout << (int)s[0] << endl;
	else
		cout << (char)((int)((s[0]-48))*10+(int)(s[1]-48)) << endl;
	return 0;
}

解法3(相似於助教的解法)

/* Programming By Linwebs */
#include<iostream>
using namespace std;
int main() {
	// ASCII 48~57 => '0'~'9' || 65~90 => 'A'~'Z'
	char s, n;
	cin >> s;
	if(s >= 65 && s <= 90)
		cout << (int)s << endl;
	else {
		cin >> n;
		cout << (char)((s-48)*10+(n-48)) << endl;
	}
	return 0;
}

助教解法

#include<iostream>
using namespace std;
int main() {
	char c, d;
	cin >> c;
	if(c >= 'A' && c <= 'Z')
		cout << (int)c << endl;
	else {
		cin >> d;
		c = (c - '0')*10 + d - '0';
		cout << c << endl;
	}
	return 0;
}
時間:2018/10/20 AM 9:27
標籤: Linwebs

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