HDU - 1027 Ignatius and the Princess II 全排列
Ignatius and the Princess II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9380 Accepted Submission(s): 5481
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?
11 8
1 2 3 4 5 6 7 9 8 11 10
将一组数据进行全排列,再将排列好的数据按从小到大排列好,如1 2 3三个数全排列可组成123 132 213 231 312 321,第3个为213。可使用STL中的next_permutation函数进行操作。
#include<stdio.h>
#include<algorithm>
using namespace std; int main()
{
int n, m;
int a[1005];
while (~scanf("%d %d", &n, &m))
{
for (int i = 1; i <= n; i++)
{
a[i] = i;
}
m--;
while (m--)
{
next_permutation(a + 1, a + n + 1);
}
printf("%d", a[1]);
for (int i = 2; i <= n; i++)
{
printf(" %d", a[i]);
}
printf("\n");
}
return 0;
}
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