Ignatius and the Princess II(全排列)

本文探讨了一段充满挑战的故事,主角伊格纳修斯面对鬼王贝尔泽布乌5166的考验,通过解决寻找特定序列的问题来拯救公主。文章深入解析了如何使用全排列思想解决此数学谜题,并提供了简洁明了的代码实现。

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 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
For each test case, you only have to output the sequence satisfied the BEelzebub's demand. When output a sequence, you should print a space between two numbers, but do not output any spaces after the last number.
 

 

Sample Input
6 4 11 8
 

 

Sample Output
1 2 3 5 6 4 1 2 3 4 5 6 7 9 8 11 10
 

题解:水题,让找第M小的序列;它这里的小应该就是逆序数从小到大;

其实就是个全排列,递归下就好了;

代码:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
const int MAXN=1010;
int ans[MAXN];
int vis[MAXN];
int N,M;
int cnt;
int flot;
void dfs(int num){
    if(flot)return;
    if(num==N){
        cnt++;
        if(cnt==M){
            for(int i=0;i<N;i++){
                if(i)P_;
                printf("%d",ans[i]);
            }
            puts("");
            flot=1;
        }
        return ;
    }
    for(int i=0;i<N;i++){
        if(vis[i+1])continue;
        ans[num]=i+1;
        vis[i+1]=1;
        dfs(num+1);
        vis[i+1]=0;
    }
}
int main(){
     while(~scanf("%d%d",&N,&M)){
         mem(vis,0);
         cnt=flot=0;
         dfs(0);
     }
    return 0;
} 

 

转载于:https://www.cnblogs.com/handsomecui/p/5244613.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值