H - What Is Your Grade?
“Point, point, life of student!” This is a ballad（歌谣）well known in colleges, and you must care about your score in this exam too. How many points can you get? Now, I told you the rules which are used in this course. There are 5 problems in this final exam. And I will give you 100 points if you can solve all 5 problems; of course, it is fairly difficulty for many of you. If you can solve 4 problems, you can also get a high score 95 or 90 (you can get the former(前者) only when your rank is in the first half of all students who solve 4 problems). Analogically（以此类推）, you can get 85、80、75、70、65、60. But you will not pass this exam if you solve nothing problem, and I will mark your score with 50. Note, only 1 student will get the score 95 when 3 students have solved 4 problems. I wish you all can pass the exam! Come on!
Input
Input contains multiple test cases. Each test case contains an integer N (1<=N<=100, the number of students) in a line first, and then N lines follow. Each line contains P (0<=P<=5 number of problems that have been solved) and T（consumed time）. You can assume that all data are different when 0<p.
A test case starting with a negative integer terminates the input and this test case should not to be processed.
Output
Output the scores of N students in N lines for each case, and there is a blank line after each case.
Sample Input
4
5 06:30:17
4 07:31:27
4 08:12:12
4 05:23:13
1
5 06:30:17
-1
Sample Output
100
90
90
95
100
易错点：
you can get the former(前者) only when your rank is in the first half of all students who solve 4 problems。即得到4分的前一半学生能得到95。以此类推。AC Code#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct Data
{
int p;
char time[10];
int t;
int grade;
int id;
};
bool cmp1(Data x, Data y)
{
if(x.p > y.p) return 1;
else if(x.p == y.p)
{
return x.t < y.t;
}
else return 0;
}
bool cmp2(Data x, Data y)
{
return x.id < y.id;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
if(n <= 0) break;
Data st[n+1];
for(int i=1; i<=n; i++)
{
scanf("%d %s",&st[i].p,st[i].time);
st[i].t = ((st[i].time[0]-'0')*10 + (st[i].time[1]-'0'))*3600 +
((st[i].time[3]-'0')*10 + (st[i].time[4]-'0'))*60 +
((st[i].time[6]-'0')*10 + (st[i].time[7]-'0'));
st[i].id = i;
}
sort(st+1,st+1+n,cmp1);
int a=0, b=0, c=0, d=0;
for(int i=1; i<=n; i++)
{
if(st[i].p == 4) a++;
if(st[i].p == 3) b++;
if(st[i].p == 2) c++;
if(st[i].p == 1) d++;
}
int x=0, y=0, z=0, m=0;
for(int i=1; i<=n; i++)
{
if(st[i].p == 5) st[i].grade = 100;
else if(st[i].p == 4)
{
if(x==0)
{
st[i].grade = 95;
x++;
}
else if(x < (a/2))
{
st[i].grade = 95;
x++;
}
else
{
st[i].grade = 90;
}
}
else if(st[i].p == 3)
{
if(y==0)
{
st[i].grade = 85;
y++;
}
else if(y < (b/2))
{
st[i].grade = 85;
y++;
}
else
{
st[i].grade = 80;
}
}
else if(st[i].p == 2)
{
if(z==0)
{
st[i].grade = 75;
z++;
}
else if(z < (c/2))
{
st[i].grade = 75;
z++;
}
else
{
st[i].grade = 70;
}
}
else if(st[i].p == 1)
{
if(m==0)
{
st[i].grade = 65;
m++;
}
else if(m < (d/2))
{
st[i].grade = 65;
m++;
}
else
{
st[i].grade = 60;
}
}
else st[i].grade = 50;
}
sort(st+1,st+1+n,cmp2);
for(int i=1; i<=n; i++)
{
printf("%d\n",st[i].grade);
}
printf("\n");
}
return 0;
}
