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PAT (Basic Level) Practice 1048 数字加密 (20 分)

作者:互联网

题目:1048 数字加密 (20 分)

来源:PAT (Basic Level) Practice

传送门 1048 数字加密

题面

image

题意:给定两个数字A,B,按给定规律对奇偶位分别操作。

思路:建议字符串输入,如果一个字符串长度小于另一个的话,前面要用零补上。

Code

点击查看代码
#include <iostream>
#include <string>
#include <cmath>
using namespace std;
string a, b;
int main() {
	bool f = false;
	cin >> a >> b;

	int ns = max(a.size(),b.size()) - min(a.size(), b.size());
	if (a.size() > b.size()) {
		for (int i = 0; i < ns; i++)
			b = '0' + b;
	}
	else if (a.size() < b.size()) {
		for (int i = 0; i < ns; i++)
			a = '0' + a;
	}
	if (a.size() % 2 == 0)f = true;
	for (int i = 0; i < b.size(); i++) {
		if (!f) {
			f = true;
			int t = a[i] - '0' + b[i] - '0';
			t = t % 13;
			if (t == 10)cout << "J";
			else if (t == 11)cout << "Q";
			else if (t == 12)cout << "K";
			else cout << t;

		}
		else if (f) {
			f = false;
			int t = b[i] - '0' - (a[i] - '0');
			if (t >= 0)cout << t;
			else cout << t + 10;

		}
	}
	cout << "\n";
	return 0;
}

标签:include,PAT,Level,int,Practice,1048,++,ns,size
来源: https://www.cnblogs.com/w0x59-h/p/15860571.html