CF周赛 B. Maximum Rounding 模拟

‍ B. Maximum Rounding

Given a natural number $x$. You can perform the following operation:

• choose a positive integer $k$ and round $x$ to the $k$-th digit

Note that the positions are numbered from right to left, starting from zero. If the number has $k$ digits, it is considered that the digit at the $k$-th position is equal to $0$.

The rounding is done as follows:

• if the digit at the $(k-1)$-th position is greater than or equal to $5$, then the digit at the $k$-th position is increased by $1$, otherwise the digit at the $k$-th position remains unchanged (mathematical rounding is used).

• if before the operations the digit at the $k$-th position was $9$, and it should be increased by $1$, then we search for the least position $k^{\prime }$ (KaTeX parse error: Expected 'EOF', got '&' at position 3: k'&̲gt;k), where the digit at the $k^{\prime }$-th position is less than $9$ and add $1$ to the digit at the $k^{\prime }$-th position. Then we assign $k=k^{\prime }$.

• after that, all digits which positions are less than $k$ are replaced with zeros.

Your task is to make $x$ as large as possible, if you can perform the operation as many times as you want.

For example, if $x$ is equal to $3451$, then if you choose consecutively:

• $k=1$, then after the operation $x$ will become $3450$
• $k=2$, then after the operation $x$ will become $3500$
• $k=3$, then after the operation $x$ will become $4000$
• $k=4$, then after the operation $x$ will become $0$

To maximize the answer, you need to choose $k=2$ first, and then $k=3$, then the number will become $4000$.

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AC code

import java.util.*;

public class Main
{
	public static void main(String[] args)
	{
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		while (T-- > 0)
		{
			String x = sc.next();
			x = "0" + x;
			int len = x.length();
			char[] s = x.toCharArray();//转为 char 数组方便操作
			int p = len;
			for (int i = len - 1; i >= 0; i--)// i 从低位开始枚举，逢 >= 5 往高位进1
			{
				if (s[i] >= '5')
				{
					s[i - 1]++;
					p = i;// 记录已经进位的最高位，表示此位以后都是 0
				}
			}
//			如果有前导零，直接跳过										
			for (int i = (s[0] == '0') ? 1 : 0; i < s.length; i++)
			{
				System.out.print(i >= p ? '0' : s[i]);//找到最近的一个进位的位置，后边的全部置零
			}
			System.out.println();
		}
	}

}