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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

原來permutation有這樣的cycle的，很好玩。

下面程式帶了個自製的輸入輸出int的處理。使用getc, putchar這些輸入輸出的確會快很多的。

#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;}