hdoj 1027 Ignatius and the Princess II 【逆康托展開】
來源：程序員人生 發布時間：2014-11-04 08:23:38 閱讀次數：2179次
Ignatius and the Princess II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4865 Accepted Submission(s): 2929
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...N⑴,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,N⑴. 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
Sample Output
1 2 3 5 6 4
1 2 3 4 5 6 7 9 8 11 10
注意：由于1000的階乘太大，而且M小于等于10000，所以我們只需要算到階乘大于10000的為就能夠了，也就是8！。以后判斷是否是第8位的特殊判斷便可。
代碼：
#include <stdio.h>
#include <string.h>
int a[9] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320};
int vis[1005];
int main(){
int n, m;
while(scanf("%d%d", &n, &m) == 2){
memset(vis, 0, sizeof(vis));
m -= 1;
int cou, temp = 1;
while(temp < n){
if((n - temp) <= 8){
int s = m/a[n-temp];
int p = m%a[n-temp];
int c = 0;
for(int i = 1; i <= n; i ++){
if(!vis[i]) ++c;
if((c⑴) == s){
printf("%d ", i);
vis[i] = 1; break;
}
}
m = p;
}
else{
for(int i = 1; i <= n; i ++){
if(!vis[i]) {
vis[i] = 1;
printf("%d ", i); break;
}
}
}
++temp;
}
for(int i = 1; i <= n; i ++){
if(!vis[i]) printf("%d
", i);
}
}
return 0;
}
生活不易，碼農辛苦
如果您覺得本網站對您的學習有所幫助,可以手機掃描二維碼進行捐贈
