CSUOJ 1270 Swap Digits

最新推荐文章于 2024-09-12 10:27:56 发布

CSU_lmw

最新推荐文章于 2024-09-12 10:27:56 发布

阅读量248

点赞数

CC 4.0 BY-SA版权

分类专栏： CSUOJ 文章标签： ACM

本文链接：https://blog.youkuaiyun.com/leo6033/article/details/80259570

CSUOJ 专栏收录该内容

39 篇文章

订阅专栏

本文介绍了一个算法问题：通过有限次交换相邻数字的位置来获得最大的可能数值。提供了C++实现代码，该代码通过寻找并移动当前可及范围内的最大数字到最左侧来实现目标。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

Now we have a number, you can swap any two adjacent digits of it, but you can not swap more than K times. Then, what is the largest probable number that we can get after your swapping?

Input

There is an integer T (1 <= T <= 200) in the first line, means there are T test cases in total.

For each test case, there is an integer K (0 <= K < 10⁶) in the first line, which has the same meaning as above. And the number is in the next line. It has at most 1000 digits, and will not start with 0.

There are at most 10 test cases that satisfy the number of digits is larger than 100.

Output

For each test case, you should print the largest probable number that we can get after your swapping.

Sample Input

Sample Output

3124
4213
4321

Hint

暴力

#include<stdio.h>
#include<string>
#include<string.h>
#include<algorithm>
#include<iostream>
typedef long long ll;
using namespace std;
int T, s, len;
char ch[1010];
int main()
{	
	cin >> T;
	while (T--)
	{
		cin >> s >> ch;
		len = strlen(ch);
		for (int i = 0; i < len; i++)
		{
			if (s <= 0)break;
			char max = '0';
			int key;
			for (int j = i + 1; j < len && j <= i + s; j++)//找到能移动的最大位数
			{
				if (max < ch[j])
				{
					max = ch[j];
					key = j;
				}
			}
			if (max > ch[i])
			{
				for (int j = key; j > i; j--)
					ch[j] = ch[j - 1];
				ch[i] = max, s =s-( key - i);
			}
		}
		cout << ch << endl;
	}
	return 0;
}
/**********************************************************************
	Problem: 1270
	User: leo6033
	Language: C++
	Result: AC
	Time:12 ms
	Memory:2024 kb
**********************************************************************/