地址：http://acm.hdu.edu.cn/showproblem.php?pid=1027

題目：

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?

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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.

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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.

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Sample Input 6 4 11 8

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Sample Output 1 2 3 5 6 4 1 2 3 4 5 6 7 9 8 11 10 思路： 通過觀察可以發現，n個數有n！種排法； 所以可以先列出1-9內的階乘，然后求出要改變多少個數的位置； 舉個例子： 9 21 t=21-1；（因為從小大大排列是最小的順序，） t/3!=3 ,不等于0，所以要改變最后3+1個數字（6 7 8 9），?推出選第3+1個數字選9；剩下（6 7 8） 令t=t%3； t/2=1;推出選第1+1個數字7；剩下（6 8）； t=t%2; t/1=0;推出選第1+0個數字6；剩下（ 8）； t=t%2; t/1=0;推出選第1+0個數字6；無剩下數字； 結束； 另外注意下輸出格式，第一次交的時候就是PE了。。。。。 1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cmath> 5 #include <cstring> 6 #include <queue> 7 #include <stack> 8 #include <map> 9 #include <vector> 10 11 #define PI acos((double)-1) 12 #define E exp(double(1)) 13 using namespace std; 14 int j[10]; 15 int num[1000]; 16 void so(int n,int m) 17 { 18 int change=9,point,all=n; 19 memset(num,0,sizeof(num)); 20 while(m/j[change]==0) 21 change--; 22 point = n-change; 23 all=change +1; 24 for(int i =point; i<=n; i++) 25 num[i]=1; 26 printf("%d",1); 27 for(int i =2; i<point; i++) 28 printf(" %d",i); 29 while(all--) 30 { 31 int t=m/j[change]; 32 for(int i =point,k=0; i<=n; i++) 33 if(num[i]) 34 { 35 if(k==t) 36 { 37 printf(" %d",i); 38 num[i]=0; 39 break; 40 } 41 k++; 42 } 43 m%=j[change--]; 44 } 45 putchar('\n'); 46 } 47 int main(void) 48 { 49 int n, m; 50 j[0]=j[1] = 1; 51 for (int i = 2; i <= 9; i++) 52 j[i] = j[i - 1] * i; 53 while (scanf("%d%d", &n, &m) == 2) 54 { 55 if(m == 1) 56 { 57 for(int i=1; i<=n-1; i++) 58 printf("%d ",i); 59 printf("%d\n",n); 60 } 61 else 62 so(n,m-1); 63 } 64 65 66 return 0; 67 } View Code

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轉載于:https://www.cnblogs.com/weeping/p/5392449.html

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