ignitius and princess 2(全排列)
A - Ignatius and the Princess II
"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?
InputThe 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.
OutputFor 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 译文:
“现在我会告诉你第一个问题。” feng5166说:“给定一个从1到N的序列,我们定义1,2,3 ... N-1,N是可以由1到N组成的所有序列中最小的序列(每个数可以在这个问题中只能使用一次)所以很容易看到第二小的序列是1,2,3 ... N,N-1现在我给你两个数字,N和M。告诉我由第1号到第N号组成的第M个最小序列。很容易,不是吗?哈哈哈哈哈......“
你能帮助伊格内修斯解决这个问题吗?
输入输入包含多个测试用例。每个测试案例由两个数字组成,N和M(1 <= N <= 1000,1 <= M <= 10000)。你可能会认为总是有一个序列满足BEelzebub的需求。输入由文件结尾终止。
产量对于每个测试用例,您只需输出满足BEelzebub需求的序列。输出序列时,应在两个数字之间打印空格,但不要在最后一个数字后面输出任何空格。
示例输入
6 4
11 8
示例输出
1 2 3 5 6 4
1 2 3 4 5 6 7 9 8 11 10
STL中的algorithm 的全排列应用
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
#define N 100010
using namespace std;
int a[N];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=;i<=n;i++)
a[i]=i;
int k=;
while(next_permutation(a+,a+n+))//全排列函数
{
k++;
if(k==m)
break;
}
for(int i=;i<n;i++)
printf("%d ",a[i]);
printf("%d\n",a[n]);
}
return ;
}
(上为博客园找的解题代码)
解题的难度在于看不懂题意,没弄懂他需要我求什么,然后就是前段时间对于算法函数algorithm ,数值算法numeric和函数对象functional的理解不够,只是理解了表层的意思,不懂灵活的应用,或者说是没有认真的应用过,经验不足;
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int N, M, a[];
while (cin >> N >> M)
{
for (int i = ; i < N; i++)
a[i] = i + ;
int mark = ;
while (next_permutation(a,a+N))
{
mark++;
if (mark == M)
break;
}
for (int i = ; i < N-; i++)
cout << a[i] << " ";
cout << a[N - ] <<endl;
}
return ;
}
(附:经过测试,如果不加控制条件用while一直循环,最后的结束next_permutation是的最后情况)
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