链接:传送门 题意:给出一个 n ,求 1 - n 全排列的第 m 个排列情况 思路:经典逆康托展开,需要注意的时要在原来逆康托展开的模板上改动一些地方. 分析:已知 1 <= M <= 10000,10000 < 8!,根据逆康托展开的原理可以发现,A[n] * (n-1)! + A[n-1] * (n-2)! + A[n-2] * (n-3)! + ...... + A[2] * 1! + A[1] * 0! ,在前 n - 8 项之前,Ai == 0,所以每次都是取剩余排列中第 0…

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…

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…

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…

#include<iostream> #include<cstdio> #include<algorithm> using namespace std; const int SIZE=1002; int main() { int n,m; int i,count; int seq[SIZE]; while(scanf("%d %d",&n,&m)!=EOF) { for(i=0;i<SIZE;i++) seq[i]=i+1; c…

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…

题目描述: 题目大意:现在给我们两个数字,N和M.我们应该编程找出由1到N组成的第M个最小序列.主要运用了全排列的思想,运用了全排列中next_permutation()函数: next_permutation 使用方法:next_permutation(数组头地址,数组尾地址);若下一个排列存在,则返回真,如果不存在则返回假 举例: 对于数组a={1,2,3}; do { cout<<a[0]<<" "<<a[1]<<" &q…

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…

题目信息:给出n.m,求n个数的按字典序排列的第m个序列 http://acm.hdu.edu.cn/showproblem.php? pid=1027 AC代码: /**  *全排列的个数(次序)  */ #include<iostream> #include<cstdio> #include<algorithm> int a[1001],x; using namespace std; void print(int n){     for(int i=1;i<n…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14475    Accepted Submission(s): 8296 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

POJ题目分类 | POJ题目分类 | HDU题目分类 | ZOJ题目分类 | SOJ题目分类 | HOJ题目分类 | FOJ题目分类 | 模拟题: POJ1006 POJ1008 POJ1013 POJ1016 POJ1017 POJ1169 POJ1298 POJ1326 POJ1350 POJ1363 POJ1676 POJ1786 POJ1791 POJ1835 POJ1970 POJ2317 POJ2325 POJ2390 POJ1012 POJ1082 POJ1099 POJ1114…

排列组合是数学中的一个分支.在计算机编程方面也有非常多的应用,主要有排列公式和组合公式.错排公式.母函数.Catalan Number(卡特兰数)等. 一.有关组合数学的公式 1.排列公式   P(n,r)=n!/r! 2.组合公式   C(n,r)=n!/(r!*(n-r)!)  C(n,r)=C(n-1,r)+C(n-1,r-1) 3.错排公式   d[1]=0;   d[2]=1; d[n]=(n-1)*(d[n-1]+d[n-2]) 4.卡特兰数 前几项:1, 2, 5, 14, 42,…

hdu1021 给n,看费波纳列数能否被3整除 算是找规律吧,以后碰到这种题就打打表找找规律吧 #include <stdio.h> int main(void) { int n; while(scanf("%d", &n) != EOF) { == || n%==) printf("yes\n"); else printf("no\n"); } ; } hdu1022 栈的简单利用吧 给两个队列  看第一个队列是否可以用第二…