Sample Input6 4 //输出第4个全排列11 8 Sample Output1 2 3 5 6 41 2 3 4 5 6 7 9 8 11 10 # include <cstdio> # include <iostream> # include <algorithm> using namespace std ; ] ; int main () { int n , m ; while (scanf("%d %d" , &n , &a…

这是一个c++函数,包含在头文件<algorithm>里面,下面是基本格式. 1 int a[]; 2 do{ 3 4 }while(next_permutation(a,a+n)); 下面的代码可产生1~n的全排列. #include <stdio.h> #include <algorithm> using namespace std; int main(){ int n; while(scanf("%d",&n)&&n){…

2018 Multi-University Training Contest 5 6351.Beautiful Now 题意就是交换任意两个数字,问你交换k次之后的最小值和最大值. 官方题解: 哇塞,C++竟然自带全排列的函数,头文件为#include<algorithm> 全排列 next_permutation() 函数.随便传送两篇博客: 1.[算法]——全排列(Permutation)以及next_permutation 2.next_permutation 代码: //1002-63…

全排列next_permutation()用法 在头文件aglorithm里 就是1~n数组的现在的字典序到最大的字典序的依次增加.(最多可以是n!种情况) int a[n]; do{ }while(next_permutation(a,a+n)); 或者知道有多少种情况 比如排好序就有n! int a[n]; sort(a,a+n); int chi=1; for(int i=1;i<=n;i++){ chi*=i; } while(chi--){ next_permutation(a,a+n…

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…

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): 9380    Accepted Submission(s): 5481 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

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…

排列2 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 11462    Accepted Submission(s): 4045 Problem Description Ray又对数字的列产生了兴趣: 现有四张卡片,用这四张卡片能排列出很多不同的4位数,要求按从小到大的顺序输出这些4位数. Input 每组数据占一行,代表四张卡片上的…