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 o…

题目链接 /* Name: Copyright: Author: Date: 2018/5/2 11:07:16 Description:输出第m小的序列 */ #include <iostream> #include <cstdio> #include <vector> #include <algorithm> #include <cstring> using namespace std; ,,,,,,,,};//阶乘 //康托展开的逆运算,{…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1027 Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10388    Accepted Submission(s): 5978 Problem Description Now our…

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 o…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9380    Accepted Submission(s): 5481 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9458    Accepted Submission(s): 5532 Problem Description Now our hero finds the door to the BEelzebub feng5166. He op…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12948    Accepted Submission(s): 7412 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

题意: 给N和M. 输出1,2,...,N的第M大全排列. 思路: 将M逆康托,求出a1,a2,...aN. 看代码. 代码: int const MAXM=10000; int fac[15]; int ans[1005]; int kk; int n,m; vector<int> pq; int main(){ int cn=0; fac[0]=1; while(1){ ++cn; fac[cn]=fac[cn-1]*cn; if(fac[cn]>MAXM){ --cn; break…

这道题目最开始完全不懂,后来百度了一下,原来是字典序.而且还是组合数学里的东西.看字典序的算法看了半天才搞清楚,自己仔细想了想,确实也是那么回事儿.对于长度为n的数组a,算法如下:(1)从右向左扫描,找到满足a[i]<a[i+1]的第一个i,也就是i = max{i|a[i]<a[i+1]},同时也意味着a[i+1]~a[n]是升序:(2)从右向左扫描,找到满足a[j]>a[i]的第一个j,也就是j = max{j|a[j]>a[i]},a[j]也是满足大于a[i]的最小数:(3)…

直接选择序列的方法解本题,可是最坏时间效率是O(n*n),故此不能达到0MS. 使用删除优化,那么就能够达到0MS了. 删除优化就是当须要删除数组中的元素为第一个元素的时候,那么就直接移动数组的头指针就能够了,那么时间效率就是O(1)了,而普通的删除那么时间效率是O(n),故此大大优化了程序. 怎样直接选择第k个序列,能够參考本博客的Leetcode题解.Leetcode题有个一模一样的题目.只是没有使用删除优化. 看见本题的讨论中基本上都是使用STL解,还有沾沾自喜的家伙,只是使用STL解决本…