PAT甲级 1153 Decode Registration Card of PAT (25 分)

最新推荐文章于 2021-09-21 05:35:50 发布

原创最新推荐文章于 2021-09-21 05:35:50 发布 · 139 阅读

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本文介绍了一个比赛成绩管理系统的设计与实现，该系统使用C++语言，通过读取选手信息和成绩，支持按等级、地点和日期查询及排序功能。系统能够处理大量数据，提供快速查询和结果输出。

题意

比较长就不解释了

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 10;
typedef long long ll;
typedef pair<int, int> pii;
#define fi first
#define se second

struct node{
	string id;
	int score, site, date;
	node(){}
	node(char *s, int a)
	{
		id = s;
		score = a;
		site = stoi(id.substr(1, 3));
		date = stoi(id.substr(4, 6));
	}
}arr[N];


bool cmp1(const node& a, const node& b)
{
	if (a.score == b.score)
		return a.id < b.id;
	else
		return a.score > b.score;
}

bool cmp2(const pii& a, const pii& b)
{
	if (a.se == b.se)
		return a.fi < b.fi;
	else
		return a.se > b.se;
}

int main(void)
{
	int i, j, temp, ca;
	int n, q;
	scanf("%d%d", &n, &q);
	char str[N];
	
	for (i = 1; i <= n; i++)
	{
		scanf("%s%d", str, &temp);
		arr[i] = node(str, temp);
	}
	for (ca = 1; ca <= q; ca++)
	{
		int op;
		scanf("%d ", &op);
		if (op == 1)
		{
			char level = getchar();
			printf("Case %d: %d %c\n", ca, op, level);
			vector<node> res;
			for (i = 1; i <= n; i++)
			{
				if (arr[i].id[0] == level)
					res.push_back(arr[i]);
			}
			sort(res.begin(), res.end(), cmp1);
			if (res.size() == 0)
				printf("NA\n");
			else
			{
				for (i = 0; i < res.size(); i++)
					printf("%s %d\n", res[i].id.c_str(), res[i].score);
			}
		}
		else if (op == 2)
		{
			scanf("%d", &temp);
			printf("Case %d: %d %03d\n", ca, op, temp);
			int num = 0, tot = 0;
			for (i = 1; i <= n; i++)
			{
				if (arr[i].site == temp)
					++num, tot += arr[i].score;
			}
			if (!num)
				printf("NA\n");
			else
			{
				printf("%d %d\n", num, tot);
			}	
		}
		else
		{
			scanf("%d", &temp);
			printf("Case %d: %d %06d\n", ca, op, temp);
			unordered_map<int, int> mp;
			vector<pii> res;
			for (i = 1; i <= n; i++)
			{
				if (arr[i].date == temp)
					++mp[arr[i].site];
			}
			for (auto it = mp.begin(); it != mp.end(); ++it)
				res.push_back(make_pair(it->fi, it->se));
			sort(res.begin(), res.end(), cmp2);
			if (res.size() == 0)
				printf("NA\n");
			else
			{
				for (i = 0; i < res.size(); i++)
					printf("%d %d\n", res[i].fi, res[i].se);
			}	
		}
	}
}