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…

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…

第 m 大的 n 个数全排列 DFS可过 #include <iostream> using namespace std; int n,m; ]; bool flag; ]; void dfs(int x) { if(x>n){ m--; ; return ; } if(!flag){ ;i<=n;i++) { if(!vis[i]&& !flag) { vis[i]=; ans[x]=i; dfs(x+); vis[i]=; } } } } int main()…

解题报告:1-n这n个数,有n!中不同的排列,将这n!个数列按照字典序排序,输出第m个数列. 第一次TLE了,没注意到题目上的n和m的范围,n的范围是小于1000的,然后m的范围是小于10000的,很明显,如果暴力枚举的话,时间复杂度就是O(n*m),为10^7,所以可以通过.这题其实就是一个排列生成的题,排列生成有多种方法,这题的关键是当找到答案的时候,要及时退出暴力过程,要不然肯定TLE. #include<cstdio> #include<iostream> #include…

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…

题意:产生第m大的排列 思路:使用 next_permutation函数(头文件algorithm) #include<iostream> #include<stdio.h> #include<algorithm> using namespace std; int main(){ ],n,m,i; while(~scanf("%d%d",&n,&m)){ ;i<=n;++i)a[i]=i; ;i<m;++i)next_pe…

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…

转载请注明出处:http://blog.csdn.net/u012860063 题目链接:http://acm.hdu.edu.cn/showproblem.php? pid=1027 Ignatius and the Princess II Problem Description Now our hero finds the door to the BEelzebub feng5166. He opens the door and finds feng5166 is about to kill…

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): 12948    Accepted Submission(s): 7412 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): 6359    Accepted Submission(s): 3760 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

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)…

链接: 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): 5822    Accepted Submission(s): 3433 Problem Description Now our h…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4571    Accepted Submission(s): 2733 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

Ignatius and the Princess II Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 2   Accepted Submission(s) : 1 Problem Description Now our hero finds the door to the BEelzebub feng5166. He opens the…

转载请注明出处:http://blog.csdn.net/lttree Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4436    Accepted Submission(s): 2642 Problem Description Now our hero finds the…

转载请注明出处:http://blog.csdn.net/lttree Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4436    Accepted Submission(s): 2642 Problem Description Now our hero finds the…

 HDU 1028 Ignatius and the Princess III Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Description "Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says. &q…

题目链接:hdu 1028 Ignatius and the Princess III 题意:对于给定的n,问有多少种组成方式 思路:dp[i][j],i表示要求的数,j表示组成i的最大值,最后答案是dp[i][i].那么dp[i][j]=dp[i][j-1]+dp[i-j][i-j],dp[i][j-1]是累加1到j-1的结果,dp[i-j][i-j]表示的就是最大为j,然后i-j有多少种表达方式啦.因为i-j可能大于j,这与我们定义的j为最大值矛盾,所以要去掉大于j的那些值 /*******…

HDU 1029 题目大意:给定数字n(n <= 999999 且n为奇数 )以及n个数,找出至少出现(n+1)/2次的数 解题思路:n个数遍历过去,可以用一个map(也可以用数组)记录每个数出现的次数, 若次数一旦达到(n+1)/2,即输出a[i] 注意能出现(n+1)/2次数的最多只有一个 /* HDU 1029 *Ignatius and the Princess IV --- dp */ #include <cstdio> #include <cstring> #in…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 51 Accepted Submission(s): 42   Problem Description Now our hero finds the door to the BEelzebub feng5166. He opens the…

HDU 1029 Ignatius and the Princess IV (思维题,排序?) Description "OK, you are not too bad, em... But you can never pass the next test." feng5166 says. "I will tell you an odd number N, and then N integers. There will be a special integer among t…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9444    Accepted Submission(s): 5524 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1026 Ignatius and the Princess I Description The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166's castle. The castle…

Ignatius and the Princess I Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he g…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=1026 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=29096#problem/D Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(…

Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10541    Accepted Submission(s): 3205Special Judge Problem Description The Princess has been abducted by the BEelzebub…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1029 Ignatius and the Princess IV Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others)Total Submission(s): 42359    Accepted Submission(s): 18467 Problem Description "OK, y…