Codeforces 1328C. Ternary Xor（贪心）

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本文分享了一个编码挑战的解决方案，通过解析输入字符串，将其拆分为两个字符串并输出，涉及字符处理与条件判断。代码使用C语言实现，展示了字符串操作的技巧。

传送门

一、题目描述

在这里插入图片描述

二、AC 代码

为了锻炼大家阅读代码的能力，本题解将不撰写算法分析说明与代码编写指导。

#include<cstdio>
#pragma warning(disable: 4996)
typedef unsigned char uchar; typedef unsigned short ushort; typedef unsigned int uint;
typedef long long ll; typedef unsigned long long ull;
const uint nmax = 5e4 + 1;
char c[nmax], * p, * q, * r, * R, a[nmax], b[nmax]; uint t, n;
int main() {
	scanf("%u", &t);
	for (uint h = 0; h < t; ++h) {
		scanf("%u", &n); getchar();
		gets_s(c, nmax); p = a, q = b; R = c + n;
		for (r = c; r != R; ++p, ++q, ++r) {
			if (*r == '0') { *p = * q = '0'; }
			else if (*r == '2') { *p = *q = '1'; }
			else { *p = '1'; *q = '0'; ++p; ++q; ++r; break; }
		}
		for (; r != R; ++p, ++q, ++r) { *p = '0'; *q = *r; }
		*p = *q = 0;
		puts(a); puts(b);
	}
	return 0;
}

#include<cstdio>
#pragma warning(disable: 4996)
typedef unsigned char uchar; typedef unsigned short ushort; typedef unsigned int uint;
typedef long long ll; typedef unsigned long long ull;
const uint nmax = 5e4 + 1;
char c[nmax], a[nmax], b[nmax]; uint t, n;
int main() {
	scanf("%u", &t);
	for (uint h = 0; h < t; ++h) {
		scanf("%u", &n); getchar();
		gets_s(c, nmax); uint i = 0;
		for (i = 0; i < n; ++i) {
			if (c[i] == '0') { a[i] = b[i] = '0'; }
			else if (c[i] == '2') { a[i] = b[i] = '1'; }
			else { a[i] = '1'; b[i] = '0'; ++i; break; }
		}
		for (; i < n; ++i) { a[i] = '0'; b[i] = c[i]; }
		a[n] = b[n] = 0;
		puts(a); puts(b);
	}
	return 0;
}