C語言反轉字符串函數reverse()

The behavior of this function template is equivalent to:

template <class BidirectionalIterator>
  void reverse (BidirectionalIterator first, BidirectionalIterator last) { while ((first!=last)&&(first!=--last)) { std::iter_swap (first,last); ++first; } }

Attention:

Bidirectional iterators to the initial and final positions of the sequence to be reversed. The range used is [first,last), which contains all the elements between first and last, including the element pointed by first but not the element pointed by last.
BidirectionalIterator shall point to a type for which swap is properly defined.

Example:

// reverse algorithm example
#include <iostream>     // std::cout
#include <algorithm>    // std::reverse
#include <vector>       // std::vector

int main () { std::vector<int> myvector; // set some values:
  for (int i=1; i<10; ++i) myvector.push_back(i);   // 1 2 3 4 5 6 7 8 9
 std::reverse(myvector.begin(),myvector.end()); // 9 8 7 6 5 4 3 2 1 // print out content:
  std::cout << "myvector contains:"; for (std::vector<int>::iterator it=myvector.begin(); it!=myvector.end(); ++it) std::cout << ' ' << *it; std::cout << '\n'; return 0; }

Output:

myvector contains: 9 8 7 6 5 4 3 2 1

實例：PAT乙級

1008 數組元素循環右移問題 (20 分)

一個數組A中存有N（>0）個整數，在不容許使用另外數組的前提下，將每一個整數循環向右移M（≥0）個位置，即將A中的數據由（A​0​​A​1​​⋯A​N−1​​）變換爲（A​N−M​​⋯A​N−1​​A​0​​A​1​​⋯A​N−M−1​​）（最後M個數循環移至最前面的M個位置）。若是須要考慮程序移動數據的次數儘可能少，要如何設計移動的方法？

輸入格式:

每一個輸入包含一個測試用例，第1行輸入N（1≤N≤100）和M（≥0）；第2行輸入N個整數，之間用空格分隔。

輸出格式:

在一行中輸出循環右移M位之後的整數序列，之間用空格分隔，序列結尾不能有多餘空格。

輸入樣例:

6 2
1 2 3 4 5 6

輸出樣例:

5 6 1 2 3 4

代碼：

#include <iostream> #include <algorithm>
using namespace std; int main(){ int n,m; int a[1000]; cin>>n>>m; m%=n; for(int i=0;i<n;i++){ cin>>a[i]; } reverse(a,a+n); reverse(a+m,a+n); reverse(a,a+m); for(int i=0;i<n;i++){ cout<<a[i]; if(i!=n-1) cout<<" "; } return 0; }