2014北京市小学生程序设计友谊赛详细答案

古城微笑少年丶 2022-04-04 13:37 179阅读 0赞

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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