[luoguP1098] 字符串的展开（模拟）

传送门

一个模拟。

代码

#include <cstdio>
#include <cstring>
#include <iostream>
#define isword(x) ((x) >= 'a' && (x) <= 'z')
#define swap(x, y) ((x) ^= (y) ^= (x) ^= (y))

int n, p1, p2, p3, add, d, start, end;
char s[1000001];

int main()
{
	int i, j, k;
	scanf("%d %d %d", &p1, &p2, &p3);
	scanf("%s", s + 1);
	n = strlen(s + 1);
	for(i = 1; i <= n; i++)
	{
		if(s[i] != '-' || i == 1 || i == n || (isword(s[i - 1]) && isdigit(s[i + 1])) || (isdigit(s[i - 1]) && isword(s[i + 1])) || (s[i] == '-' && (s[i - 1] == '-' || s[i + 1] == '-'))) putchar(s[i]);
		else
		{
			if(s[i - 1] + 1 == s[i + 1]) continue;
			if(s[i - 1] >= s[i + 1])
			{
				putchar('-');
				continue;
			}
			add = 0;
			d = 1;
			start = s[i - 1] + 1;
			end = s[i + 1];
			if(p3 == 2) swap(start, end), start -= 1, end -= 1, d = -1;
			if(p1 == 3)
			{
				for(j = 1; j <= p2; j++)
					for(k = start; k != end; k += d) putchar('*');
				continue;
			}
			if(isword(s[i - 1]) && p1 == 2) add -= 32;
			for(k = start; k != end; k += d)
				for(j = 1; j <= p2; j++)
					putchar(k + add);
		}
	}
	return 0;
}