IOS-本地存储-NSUserDefaults
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; }
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