C语言反转字符串函数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 << '
';

  return 0;
}

Output:

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

实例：PAT乙级

1008 数组元素循环右移问题 (20 分)

一个数组A中存有N（>0）个整数，在不允许使用另外数组的前提下，将每个整数循环向右移M（≥0）个位置，即将A中的数据由（A​0​​A​1​​⋯A​N−1​​）变换为（A​N−M​​⋯A​N−1​​A​0​​A​1​​⋯A​N−M−1​​）（最后M个数循环移至最前面的M个位置）。如果需要考虑程序移动数据的次数尽量少，要如何设计移动的方法？

输入格式:

每个输入包含一个测试用例，第1行输入N（1≤N≤100）和M（≥0）；第2行输入N个整数，之间用空格分隔。

输出格式:

在一行中输出循环右移M位以后的整数序列，之间用空格分隔，序列结尾不能有多余空格。

输入样例:

6 2
1 2 3 4 5 6

输出样例:

5 6 1 2 3 4

代码：

#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;
}