2018北京小学生程序设计友谊赛详细答案

逃离我推掉我的手 2022-04-10 12:13 179阅读 0赞

# 第1题 找平方数 #

#include <iostream>
    #include <cmath>
    using namespace std;
    
    int main()
    {
        int n;
        cin >> n;
        int a[n];
        for(int i = 0; i < n; i++)
        {
            cin >> a[i];
            int root = sqrt(a[i]);
            if(root * root == a[i])
            {
                cout << a[i] << endl;
            }
        }
    
        return 0;
    }

# 第2题 回文密码 #

#include <iostream>
    using namespace std;
    
    bool isPalin(string s)
    {
        int len = s.size();
        int i = 0;
        int j = len - 1;
        while(i < j)
        {
            if(s[i] != s[j])
            {
                return false;
            }
            i++;
            j--;
        }
    
        return true;
    }
    
    int main()
    {
        string s;
    cin >> s;
    
        int len = s.size();
        int i = 0;
        if(isPalin(s))
        {
            i = 1;
        }
    
        while(i < len)
        {
            cout << s[i];
            i += 2;
        }
        cout << endl;
    
        return 0;
    }

# 第3题 采购奖品 #

#include<iostream>
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    
    struct goods
    {
    	int cost;
    	int amount;
    };
    
    bool cmp(goods a,goods b)
    {
    	return a.cost<b.cost;
    }
    
    int main()
    {
    	freopen("prize.in","r",stdin);
    	freopen("prize.out","w",stdout);
    
    	int money, n;
    	cin >> money >> n;
    
    	goods g[n];
    	int i = 0;
    	for(; i < n; i++)
    	{
    		cin >> g[i].cost >> g[i].amount;
    	}
    
    	sort(g, g + n, cmp);
    
    	int ans = 0;
    	int cnt = 0;
    
    	i = 0;
    	while(money >= g[i].cost) // 若剩余的钱连一个都不买不了，则循环结束
    	{
    		int tmpCnt = money / g[i].cost; // 不考虑商品数量限制，理论上最多可买的数量
    		cnt = min(tmpCnt, g[i].amount);
    		money -= (cnt*g[i].cost);
    		ans += cnt;
    		i++;
    	}
    
    	cout << ans << endl;
    
    	return 0;
    }

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