hdu——poj2470—— Ambiguous permutations
来源:互联网 发布:局域网控制电脑软件 编辑:程序博客网 时间:2024/05/17 00:03
Problem Description
Some programming contest problems are really tricky: not only do they require a different output format from what you might have expected, but also the sample output does not show the difference. For an example, let us look at permutations.
A permutation of the integers 1 to n is an ordering of these integers. So the natural way to represent a permutation is to list the integers in this order. With n = 5, a permutation might look like 2, 3, 4, 5, 1.
However, there is another possibility of representing a permutation: You create a list of numbers where the i-th number is the position of the integer i in the permutation. Let us call this second possibility an inverse permutation. The inverse permutation for the sequence above is 5, 1, 2, 3, 4.
An ambiguous permutation is a permutation which cannot be distinguished from its inverse permutation. The permutation 1, 4, 3, 2 for example is ambiguous, because its inverse permutation is the same. To get rid of such annoying sample test cases, you have to write a program which detects if a given permutation is ambiguous or not.
Input
The input contains several test cases.
The first line of each test case contains an integer n (1 <= n <= 100000). Then a permutation of the integers 1 to n follows in the next line. There is exactly one space character between consecutive integers. You can assume that every integer between 1 and n appears exactly once in the permutation.
The last test case is followed by a zero.
Output
For each test case output whether the permutation is ambiguous or not. Adhere to the format shown in the sample output.
Sample Input
4
1 4 3 2
5
2 3 4 5 1
1
1
0
Sample Output
ambiguous
not ambiguous
ambiguous
Some programming contest problems are really tricky: not only do they require a different output format from what you might have expected, but also the sample output does not show the difference. For an example, let us look at permutations.
A permutation of the integers 1 to n is an ordering of these integers. So the natural way to represent a permutation is to list the integers in this order. With n = 5, a permutation might look like 2, 3, 4, 5, 1.
However, there is another possibility of representing a permutation: You create a list of numbers where the i-th number is the position of the integer i in the permutation. Let us call this second possibility an inverse permutation. The inverse permutation for the sequence above is 5, 1, 2, 3, 4.
An ambiguous permutation is a permutation which cannot be distinguished from its inverse permutation. The permutation 1, 4, 3, 2 for example is ambiguous, because its inverse permutation is the same. To get rid of such annoying sample test cases, you have to write a program which detects if a given permutation is ambiguous or not.
Input
The input contains several test cases.
The first line of each test case contains an integer n (1 <= n <= 100000). Then a permutation of the integers 1 to n follows in the next line. There is exactly one space character between consecutive integers. You can assume that every integer between 1 and n appears exactly once in the permutation.
The last test case is followed by a zero.
Output
For each test case output whether the permutation is ambiguous or not. Adhere to the format shown in the sample output.
Sample Input
4
1 4 3 2
5
2 3 4 5 1
1
1
0
Sample Output
ambiguous
not ambiguous
ambiguous
#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAX = 100005;int a[MAX];int main(){ int n, i, flag; while (scanf("%d", &n) != EOF && n) { for (i = 1; i <= n; ++i) { scanf("%d", &a[i]); } flag = 1; for (i = 1; i <= n; ++i) { if (i != a[a[i]]) { flag = 0; break; } } if (flag) { printf("ambiguous\n"); } else { printf("not ambiguous\n"); } } return 0;}
- hdu——poj2470—— Ambiguous permutations
- Ambiguous Permutations
- Ambiguous permutations
- Permutations——LeetCode
- LeetCode——Permutations
- leetcode——Permutations
- LeetCode46——Permutations
- Leetcode46——Permutations
- Algorithms—46.Permutations
- poj2470
- poj2470
- poj2470
- zoj 2795 Ambiguous permutations
- poj 2470 Ambiguous permutations
- zoj 2795 Ambiguous permutations
- poj 2470 Ambiguous permutations
- ZOJ 2795 Ambiguous permutations
- ZOJ 2795 Ambiguous permutations
- IOS开发中常用到的宏定义
- hdu1501
- Linux grep命令
- 删除Oracle数据库时常见问题(注册表方面,文件目录方面,环境变量方面)
- Linux使用C读取文件目录
- hdu——poj2470—— Ambiguous permutations
- hdu 1013
- 动手写简单的嵌入式操作系统一
- DC_CheckWatchdog
- java中的BigInteger
- ARM指令集与Thumb指令集的区别
- struts学习(3)——struts核心思想
- 编程之美 1.1 让cpu占用率曲线听你指挥(多核处理器)
- modem_hotplug 脚本解析