We consider permutations of the numbers 1,..., N
for some N. By permutation we mean a rearrangment of the
number 1,...,N. For example
2 4 5 1 7 6 3 8
is a permutation of 1,2,...,8. Of course,
1 2 3 4 5 6 7 8
is also a permutation of 1,2,...,8.
We can "walk around" a permutation in a interesting way and here
is how it is done for the permutation above:
Start at position 1. At position 1 we have 2 and so we go to
position 2. Here we find 4 and so we go to position 4. Here we find
1, which is a position that we have already visited. This completes
the first part of our walk and we denote this walk by (1 2 4 1). Such
a walk is called a cycle. An interesting property of such
walks, that you may take for granted, is that the position we revisit
will always be the one we started from!
Examples
Sample input 1:
8
2 4 5 1 7 6 3 8
Sample output 1:
4
1 2 4 1
3 5 7 3
6 6
8 8
Sample input 2:
8
1 2 3 4 5 6 7 8
Sample output 2:
8
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
原来permutation有这样的cycle的，很好玩。
下面程序带了个自制的输入输出int的处理。使用getc, putchar这些输入输出的确会快很多的。
#pragma once
#include <vector>
#include <string>
#include <algorithm>
#include <stack>
#include <stdio.h>
#include <iostream>
using namespace std;
namespace
{
int scanInt2()
{
char c = getc(stdin);
while (c < ‘0‘ || c > ‘9‘)
{
c = getc(stdin);
}
int num = 0;
while (‘0‘ <= c && c <= ‘9‘)
{
num = (num<<3) + (num<<1) + c - ‘0‘;
c = getc(stdin);
}
return num;
}
void printInt(int n)
{
int i = 0;
char buffer[32] = {0};
while (n)
{
buffer[i++] = n % 10 + ‘0‘;
n /= 10;
}
//buffer[i] = ‘\0‘;
//fputs(buffer, stdout);
for (i--; i >= 0; i--)
{
fputc(buffer[i], stdout);
}
}
}
int PermutaionCycles()
{
const int n = scanInt2();
int *A = new int[n+1];
for (int i = 1; i <= n; i++)
{
A[i] = scanInt2();
}
vector<vector<int> > rs;
vector<int> tmp;
for (int i = 1; i <= n; i++)
{
if (A[i] != -1)
{
tmp.clear();
int j = i;
while (tmp.size() < 2 || tmp.back() != i)
{
tmp.push_back(j);
int t = j;
j = A[j];
A[t] = -1;
}
rs.push_back(tmp);
}
}
printInt(rs.size());
putc(‘\n‘, stdout);
for (int i = 0; i < (int)rs.size(); i++)
{
for (int j = 0; j < int(rs[i].size()); j++)
{
printInt(rs[i][j]);
putc(‘ ‘, stdout);
}
putchar(‘\n‘);
}
delete [] A;
return 0;
}
codechef Permutation Cycles 题解,布布扣,bubuko.com