HDU1027（Ignatius and the Princess II）
阿新 • • 發佈：2018-12-12
Ignatius and the Princess II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5830 Accepted Submission(s): 3439
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?
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
Sample Input 6 4 11 8
題意：給定n，若1,2,3,...n,為最小序列，確定第m小的排序序列並輸出。
思路：一個比較笨的方法，找規律。注意1<=M<=10000。
可以發現，一共是n!個序列，那麼對於後now位，其共有now！個序列，然後按照這個規律，第一次找到now！>= m，確定p值，然後累加（now - 1）！，直到計數count + k*(now - 1)! >=m，確定k值，然後找在後p位中，第k大的值，這個值即是從後往前數第p位的值，依次類推，直到確定最後一個值。
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int rec[15], tmp[15], INF = 1<<28;
int main(){
int n, m;
while(~scanf("%d %d",&n, &m)){
int t = 1, i, now, cur = 0;
for(i = 1; i <= 10; i++){
t *= i;
tmp[i] = t;
rec[i] = n - i + 1;
if(t >= m)break;
}
now = i;
sort(rec+1, rec+now+1);
for(i = 1; i <= n - now; i++)printf("%d ",i);
while(now > 1){
int temp = cur;
while(cur+tmp[now-1] < m){
cur += tmp[now-1];
}
int _ = (cur-temp)/tmp[now-1]+1;
printf("%d ",rec[_]);
rec[_] = INF;
sort(rec+1, rec+now+1);
now--;
}
printf("%d\n",rec[1]);
}
return 0;
}