next_permutation函数(全排列生成利器)

题目大意：给出一个n和一个m，要求求出数列1,2,3……n

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

6 411 8

Sample Output

1 2 3 5 6 41 2 3 4 5 6 7 9 8 11 10

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;int main(){    int n, m;    int a[1010];    int flag;    while(scanf("%d %d", &n, &m) != EOF)    {        flag = 0;        for(int i = 1; i <= n; i ++)            a[i] = i;        int cnt = 0;        do        {            cnt ++;            if(cnt == m)            {                for(int i = 1; i <= n; i ++)                    {                        if(flag)                            printf(" ");                        flag ++;                        printf("%d", a[i]);                    }                printf("\n");                break;            }        }        while(next_permutation(a + 1, a + n + 1));    }    return 0;}

next_permutation()是按照字典序产生排列的，并且是从数组中当前的字典序开始依次增大直至到最大字典序。

虽然最后一个排列没有下一个排列，用next_permutation会返回false，但是使用了这个方法后，序列会变成字典序列的第一个

hdu 1716

Ray又对数字的列产生了兴趣：   现有四张卡片，用这四张卡片能排列出很多不同的4位数，要求按从小到大的顺序输出这些4位数。   
Input
每组数据占一行，代表四张卡片上的数字（0<=数字<=9），如果四张卡片都是0，则输入结束。  
Output
对每组卡片按从小到大的顺序输出所有能由这四张卡片组成的4位数，千位数字相同的在同一行，同一行中每个四位数间用空格分隔。  每组输出数据间空一行，最后一组数据后面没有空行。  
Sample Input
1 2 3 41 1 2 30 1 2 30 0 0 0
Sample Output
1234 1243 1324 1342 1423 14322134 2143 2314 2341 2413 24313124 3142 3214 3241 3412 34214123 4132 4213 4231 4312 43211123 1132 1213 1231 1312 13212113 2131 23113112 3121 32111023 1032 1203 1230 1302 13202013 2031 2103 2130 2301 23103012 3021 3102 3120 3201 3210
#include <iostream>#include <cstdio>#include <algorithm>using namespace std;int main(){    int a[5];    int flag1 = 0;    while(scanf("%d %d %d %d", &a[1], &a[2], &a[3], &a[4]) != EOF)    {        if(a[1] == 0&&a[2] == 0&&a[3] == 0&&a[4] == 0)            break;            if(flag1)        printf("\n");        flag1 ++;            int flag = 0;            int j = a[1];        do        {            if(a[1] == 0)                continue;                if(a[1] != j&&j != 0)                    {                        printf("\n");                        flag = 0;                    }                int tem = a[1]*10*10*10 + a[2]*10*10 +a[3]*10 + a[4];                if(flag)                    printf(" ");                    flag ++;                printf("%d",tem);                j = a[1];        }while(next_permutation(a + 1, a + 5));        printf("\n");    }    return 0;}