Swap Digits

最新推荐文章于 2024-09-12 10:27:56 发布

JakeRame

最新推荐文章于 2024-09-12 10:27:56 发布

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分类专栏：贪心

本文链接：https://blog.youkuaiyun.com/JakeRame/article/details/80552076

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本文介绍了一个算法问题，即如何通过有限次相邻数字交换得到最大可能的整数。该问题适用于最多1000位的数字，并限制了交换次数。文章提供了一种贪心算法的解决方案，通过将高位尽可能置为较大数值的方式逐步构建最大数。

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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

思路：贪心

首先尽量把最高位变成最大的数字，再往右一位位一样的操作，直到变换次数用完

代码如下：

#include<iostream>
#include<string.h>
using namespace std;

int main()
{
	int T, s, len;
	char ch[1001];
	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++){
				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 -= key - i;
			}
		}
		cout << ch << endl;
	}
	return 0;
}