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  • 【YBTOJ】生日相同

    题目大意:

    给你几个人的名字和他们的生日,求生日相同的人的生日和他们的名字。

    正文:

    对于每个生日开一个 ( exttt{vector}),我开到了 ( exttt{1231}),如果足够聪明可能可以只开到 ( exttt{366})

    代码:

    
    inline ll READ()
    {
    	ll x = 0, f = 1;
    	char c = getchar();
    	while (c != '-' && (c < '0' || c > '9')) c = getchar();
    	if (c == '-') f = -f, c = getchar();
    	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
    	return x * f;
    }
    
    int n;
    bool flag = 1;
    vector <string> a[1300];
    
    bool cmp(string a, string b)
    {
    	int as = a.size(), bs = b.size();
    	if (as == bs) return a < b;
    	return as < bs;
    }
    
    int main()
    {
    	n = READ();
    	for (int i = 1; i <= n; i++)
    	{
    		string s; int x;
    		cin >> s;
    		x = READ() * 100 + READ();
    		if (!a[x].empty()) flag = 0;
    		a[x].push_back(s); 
    	}
    	if (flag) {puts("None"); return 0;} 
    	for (int i = 101; i <= 1231; i++)
    		if (!a[i].empty())
    		{
    			printf ("%d %d ", i / 100, i % 100);
    			sort (a[i].begin(), a[i].end(), cmp);
    			for (int j = 0; j < a[i].size(); j++)
    				cout << a[i][j] << ' ';
    			putchar(10);
    		}
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/GJY-JURUO/p/14717691.html
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