HDU 1027 组合数学

最新推荐文章于 2018-08-30 20:13:00 发布

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#组合数学

本博客通过解决一个复杂的排列问题，展示了如何通过编程帮助英雄拯救公主的故事。包括输入和输出规范，以及详细的代码实现步骤，适用于算法与数据结构的学习。

题目：

题意需要理解一下，到底怎么排序，然后就简单了

Ignatius and the Princess II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4066 Accepted Submission(s): 2436

Problem Description

Now our hero finds the door to the BEelzebub feng5166. He opens the door and finds feng5166 is about to kill our pretty Princess. But now the BEelzebub has to beat our hero first. feng5166 says, "I have three question for you, if you can work them out, I will release the Princess, or you will be my dinner, too." Ignatius says confidently, "OK, at last, I will save the Princess."

"Now I will show you the first problem." feng5166 says, "Given a sequence of number 1 to N, we define that 1,2,3...N-1,N is the smallest sequence among all the sequence which can be composed with number 1 to N(each number can be and should be use only once in this problem). So it's easy to see the second smallest sequence is 1,2,3...N,N-1. Now I will give you two numbers, N and M. You should tell me the Mth smallest sequence which is composed with number 1 to N. It's easy, isn't is? Hahahahaha......"
Can you help Ignatius to solve this problem?

Input

The input contains several test cases. Each test case consists of two numbers, N and M(1<=N<=1000, 1<=M<=10000). You may assume that there is always a sequence satisfied the BEelzebub's demand. The input is terminated by the end of file.

Output

For each test case, you only have to output the sequence satisfied the BEelzebub's demand. When output a sequence, you should print a space between two numbers, but do not output any spaces after the last number.

Sample Input

6 4
11 8

Sample Output

1 2 3 5 6 4
1 2 3 4 5 6 7 9 8 11 10

代码：

#include<stdio.h>
#include<stdlib.h>
int comp(const void*a,const void*b)
{
    return *(int*)b-*(int*)a;
}
void main()
{
	int i,j,k,n,m,a[100001],temp;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<n;i++)//数组赋初值
			a[i]=n-i;
		for(i=0;i<m-1;i++)//操作次数
		{
			for(j=0;j<n;j++)
			{
				if(a[j+1]<a[j])
					break;
			}//确定调换的位置
			for(k=0;k<n;k++)
			{
				if(a[k]>a[j+1])
				{
					temp=a[j+1];
					a[j+1]=a[k];
					a[k]=temp;
					break;
				}
			}
			qsort(a,j+1,sizeof(int),comp);//尾数排序
		}
		for(j=n-1;j>0;j--)
		{
			printf("%d ",a[j]);
		}
		printf("%d\n",a[0]);
	}
}