2014北京市小学生程序设计友谊赛详细答案

第1题

#include <iostream>
using namespace std; 

int main() 
{
  freopen("apple.in", "r", stdin);
  freopen("apple.out", "w", stdout);
  
  int a, b;
  cin >> a >> b;
  cout << (a % b ? (a / b + 1) : a / b) << endl;
  
  return 0;
}

第2题

#include <iostream>
using namespace std; 

int main() 
{
//  freopen("height.in", "r", stdin);
//  freopen("height.out", "w", stdout);
  
  int n;
  cin >> n;
  int mx = 0, secMx = 0;
  int mxPos = 0, secPos = 0;
  int a[n + 1];
  for(int i = 1; i <= n; i++)
  {
    cin >> a[i];
    if(a[i] > mx)
    {
      secMx = mx;
      secPos = mxPos;
      mx = a[i];
      mxPos = i;
    }
    else if(a[i] > secMx)
    {
      secMx = a[i];
      secPos = i;
    }
  }
  
  cout << secPos << ' ' << secMx << endl;
  return 0;
}

第3题

#include <iostream>
using namespace std; 

int main() 
{
//  freopen("change.in", "r", stdin);
//  freopen("change.out", "w", stdout);
  
  int n;
  cin >> n;
  int change = 100 - n;
  int cnt = 0;
  
  cnt += change / 50;
  change %= 50;
  
  cnt += change / 20;
  change %= 20;
  
  cnt += change / 10;
  change %= 10;
  
  cnt += change / 5;
  change %= 5;
  
  // 最后加上一元钱的数量 
  cnt += change;
  
  cout << cnt << endl; 
   
  return 0;
}

扩展：假如有三种面值纸币：1元，5元，11元（现实中没有11元的币值，但是不要纠结，只是做个假设而已）。找零15元，至少需要几张纸币？