10.6 体育成绩统计

10.6 模拟赛

T1 （LOJ 2404 体育成绩统计）

题意很简单 模拟就行了

注意事项 ：

1. 写函数把字符串化成秒， 一定要看清字符串的格式和特点， 比如说前面0是一位， 最后没有字符之类的
2. 比较和判分的函数 一定记得返回0；（表示死在这里）
3. 边读入边算是最好的解决方案， 存起来会很麻烦
4. if else 可换为数组存(无所谓)
5. 调试要有耐心， （可以把给的大样例里某个人扒出来）
6. 从主函数开始一步一步调试， 这样才有更大的一次成功的把握
7. 面对这些读入题 我们选择scanf格式读入（考场上忘了这个骚操作， 还傻傻的处理字符窜。。。）
8. 单双引号读入输出处理' ' " "

丑陋的代码（调了多半天， 所以码风怪）

#include <bits/stdc++.h>
using namespace std;
int n, m;
map<string, int> mp;
struct zz
{
	string id1;// 学号 
	int xb;    //  性别 
	int sc1;    // 体育课专项成绩 
	int s2;   //期末长跑测试成绩
	int s3;    // 体质有没有通过
	int s4;    // 大一专项计划期末成绩
	int s5;    //  参加班级训练营的次数 
	int s6;     // 阳光长跑次数 
	int last;
	int sum;
}f[4005];
int solve(string x)
{
	int i = 0, ans1 = 0,ans2 = 0;
	for(; i < x.size(); ++ i)
	{
		if(x[i] < '0' || x[i] > '9') break;
		ans1 = ans1 * 10 + x[i] - '0';
	}
	++ i;
	for(;i < x.size(); ++ i)
	{
	    if(x[i] < '0' || x[i] > '9') break;
		ans2 = ans2 * 10 + x[i] - '0' ;
	}
	return ans1 * 60 + ans2;
}
int ss(string x)
{
	int i = 0, ans1 = 0, ans2 = 0, ans3 = 0;
	for(; i < x.size(); i ++)
	{
		if(x[i] < '0' || x[i] > '9') break;
		ans1 = ans1 * 10 + x[i] - '0';
	}
	i ++;
	for(; i < x.size(); i ++)
	{
		if(x[i] < '0' || x[i] > '9') break;
		ans2 = ans2 * 10 + x[i] - '0';		
	}
	i ++;
	for(; i < x.size(); i ++)
    	ans3 = ans3 * 10 + x[i] - '0';
	return ans1 * 3600 + ans2 * 60 + ans3;
}
int ti(int a,int b){return a*60+b;}
int emm(int x)
{
	if(f[x].xb == 1)
	{
		if(f[x].s2 <= 12 *60 + 30) return 20;
		else if(f[x].s2 <= 13 *60) return 18;
		else if(f[x].s2 <= 13 *60 + 30) return 16;
		else if(f[x].s2 <= 14 * 60) return 14;
		else if(f[x].s2 <= 14 * 60 + 30) return 12;
		else if(f[x].s2 <= 15 * 60 + 10) return 10;
		else if(f[x].s2 <= 15 *60 + 50) return 8;
		else if(f[x].s2 <= 16 * 60 + 30) return 6;
		else if(f[x].s2 <= 17 *60 + 10) return 4;
		else if(f[x].s2 <= 18 * 60) return 2;

 	}
 	else
	{
		if(f[x].s2 <= 6 *60 + 40) return 20;
		else if(f[x].s2 <= 6 *60 + 57) return 18;
		else if(f[x].s2 <= 7 *60 + 14) return 16;
		else if(f[x].s2 <= 7 * 60 + 31) return 14;
		else if(f[x].s2 <= 7 * 60 + 50) return 12;
		else if(f[x].s2 <= 8 * 60 + 5) return 10;
		else if(f[x].s2 <= 8 *60 + 20) return 8;
		else if(f[x].s2 <= 8 * 60 + 35) return 6;
		else if(f[x].s2 <= 8 *60 + 50) return 4;
		else if(f[x].s2 <= 9 * 60) return 2;
 	}
 	return 0;
}
int emm2(int x)
{
	if(f[x].s6 >= 21) return 10;
	else if(f[x].s6 >= 19) return 9;
	else if(f[x].s6 >= 17) return 8;
	else if(f[x].s6 >= 14) return 7;
	else if(f[x].s6 >= 11) return 6;
	else if(f[x].s6 >= 7) return 4;
	else if(f[x].s6 >= 3) return 2;
	return 0;
}
int emm3(int x)
{
	int o = f[x].s5 + f[x].s6;
//	cout << o <<endl;
	if(o >= 18) return 5;
	else if(o >= 15) return 4;
	else if(o >= 12) return 3;
	else if(o >= 9) return  2;
	else if(o >= 6) return 1;
	return 0;
}
char tmp[10];
string  tmp1;
int tot = 0;
struct yy
{
	int date;
	string End;	
}mm[150005];
string sta, End, id;
double l;
string temp;
int step, date;
double ll;
int cnt;
signed main()
{
	freopen("a.in", "r", stdin);
	freopen("a.out", "w", stdout);
	scanf("%d", &n);
	for(int i = 1; i <= n; i ++)
	{
		cin >> f[i].id1;  
		mp[f[i].id1] = i;
		scanf("%s", tmp + 1);  f[i].xb = (tmp[1] == 'M'? 1 : 2); //1是男的 
		scanf("%d", &f[i].sc1);
		cin >> tmp1; f[i].s2 = solve(tmp1);
		scanf("%s", tmp + 1);   f[i].s3 = (tmp[1] == 'P'? 10 : 0); //1是通过
		scanf("%d", &f[i].s4);
		scanf("%d", &f[i].s5); 
	}
	scanf("%d",&m);
	for(int i = 1; i <= m; i ++)
	{
		cin >> date;
		cin >> id;
		cin >> sta;
		cin >> End;
	    cin >> l; 
		ll = l * 1000;
		cin >> temp;
		cin >> step;
		
		int x = mp[id];
		if((f[x].xb == 1&& ll < 3000.0)||(f[x].xb == 2&&ll < 1500.0)) continue;
		double t = ss(End) - ss(sta);
		double v = ll / (double)t;
//		cout << v  << endl;
		if(v < (double)2.0 || v > (double) 5.0 ) continue;
//		cout << solve(temp) <<endl;
		if(solve(temp) > 270) continue;
		
		double s = (double)ll / (double)step ;
//		cout << s <<endl;
		if(s > 1.5) continue;
		int o = f[x].last; 
		if(o == 0){
			mm[++tot].date = date; mm[tot].End = End;
			f[x].last = tot;
			f[x].s6 ++;
			continue;
		}
        if(mm[o].date == date &&ss(sta) - ss(mm[o].End) < 6 * 3600) continue;
        if(date - mm[o].date <= 1 && ss(sta) + 24 * 3600 - ss(mm[o].End) < 6 * 3600) continue;
        mm[++tot].date = date;
		mm[tot].End = End;
        f[x].last = tot;
        f[x].s6 ++;
	}
	for(int i = 1; i <= n; i ++)
	{
		f[i].sum += f[i].sc1;
		f[i].sum += emm(i);
		f[i].sum += emm2(i);
		f[i].sum += f[i].s3;
		f[i].sum += f[i].s4;
		f[i].sum += emm3(i);
	}
	for(int i = 1; i <= n; i ++)
	{
		cout << f[i].id1 << " " << f[i].sum <<" ";
//		cout << f[i].sc1 << " " << emm(i) << " " << emm2(i) << " " <<  f[i].s3 << " " << f[i].s4 << " " << emm3(i)<<endl;;
//         cout <<f[i].s5 + f[i].s6 <<endl;
		if(f[i].sum >= 95) {cout<<"A"<<"
";continue;}
			if(f[i].sum >= 90) {
			cout<<"A-"<<"
";
			continue;
		}
			if(f[i].sum >= 85) {
			cout<<"B+"<<"
";
			continue;
		}
			if(f[i].sum >= 80) {
			cout<<"B"<<"
";
			continue;
		}
			if(f[i].sum >= 77) {
			cout<<"B-"<<"
";
			continue;
		}
			if(f[i].sum >= 73) {
			cout<<"C+"<<"
";
			continue;
		}
			if(f[i].sum >= 70) {
			cout<<"C"<<"
";
			continue;
		}
			if(f[i].sum >= 67) {
			cout<<"C-"<<"
";
			continue;
		}
			if(f[i].sum >= 63) {
			cout<<"D+"<<"
";
			continue;
		}
			if(f[i].sum >= 60) {
			cout<<"D"<<"
";
			continue;
		}
		else{
			cout<<"F"<<"
";
			continue;
		}
	}
	return 0;
}