CSUOJ 1270 Swap Digits_swap digits acm-优快云博客

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

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。

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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
**********************************************************************/