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 ：
- 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
- 比较和判分的函数 一定记得返回0；（表示死在这里）
- 边读入边算是最好的解决方案， 存起来会很麻烦
- if else 可换为数组存(无所谓)
- 调试要有耐心， （可以把给的大样例里某个人扒出来）
- 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
- 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。。。）
- 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"<<"\n";continue;}
if(f[i].sum >= 90) {
cout<<"A-"<<"\n";
continue;
}
if(f[i].sum >= 85) {
cout<<"B+"<<"\n";
continue;
}
if(f[i].sum >= 80) {
cout<<"B"<<"\n";
continue;
}
if(f[i].sum >= 77) {
cout<<"B-"<<"\n";
continue;
}
if(f[i].sum >= 73) {
cout<<"C+"<<"\n";
continue;
}
if(f[i].sum >= 70) {
cout<<"C"<<"\n";
continue;
}
if(f[i].sum >= 67) {
cout<<"C-"<<"\n";
continue;
}
if(f[i].sum >= 63) {
cout<<"D+"<<"\n";
continue;
}
if(f[i].sum >= 60) {
cout<<"D"<<"\n";
continue;
}
else{
cout<<"F"<<"\n";
continue;
}
}
return 0;
}
