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？

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