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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