hdu 1027 Ignatius and the Princess II(STL全排列)
2016-10-15 16:08
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Ignatius and the Princess II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7117 Accepted Submission(s): 4228
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
题意:给n和m求出1~n的所有排列中字典序第m小的
思路:
next_permutation()函数功能是输出所有比当前排列大的排列,顺序是从小到大。
prev_permutation()函数功能是输出所有比当前排列小的排列,顺序是从大到小。
m只有10000,不会超时。
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int num[10050];
int main()
{
int n,m;
while(~scanf("%d %d",&n,&m))
{
for(int i=1;i<=n;i++)
num[i]=i;
for(int i=1;i<m;i++)
next_permutation(num+1,num+1+n);
for(int i=1;i<n;i++)
printf("%d ",num[i]);
printf("%d\n",num
);
}
return 0;
}
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