I - I
Description
“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
#include <iostream>
using namespace std;
int main ()
{
int n,i,j;
char b[110];
while(cin >> n)
{
long long a[110]={0},c[110]={0},e[110]={0};
int q[5]={0};
if(n==-1)
{
break;
}
for(i=0;i<n;++i)
{
cin >> c[i];
if(c[i]==4)
{
q[4]++;
}
if(c[i]==3)
{
q[3]++;
}
if(c[i]==2)
{
q[2]++;
}
if(c[i]==1)
{
q[1]++;
}
cin >> b;
for(j=0;b[j]!='\0';++j)
{
if(b[j]!=':')
{
a[i]+=b[j]-'0';
}
a[i]*=10;
}
}
q[4]/=2;
q[3]/=2;
q[2]/=2;
q[1]/=2;
for(i=0;i<n;++i)
{
for(j=0;j<n;++j)
{
if(c[i]==4)
{
if(a[i]>a[j]&&c[j]==4)
{
e[i]++;
}
}
else if(c[i]==3)
{
if(a[i]>a[j]&&c[j]==3)
{
e[i]++;
}
}
else if(c[i]==2)
{
if(a[i]>a[j]&&c[j]==2)
{
e[i]++;
}
}
else if(c[i]==1)
{
if(a[i]>a[j]&&c[j]==1)
{
e[i]++;
}
}
}
}
/* for(i=0;i<n;i++)
{
cout << e[i]+1 << endl;
}*/
for(i=0;i<n;i++)
{
if(c[i]==5)
{
cout << 100 << endl;
}
else if(c[i]==0)
{
cout << 50 << endl;
}
else if(c[i]==4)
{
if(e[i]+1>q[4])
{
cout << 90 << endl;
}
else
{
cout << 95 << endl;
}
}
else if(c[i]==3)
{
if(e[i]+1>q[3])
{
cout << 80 << endl;
}
else
{
cout << 85 << endl;
}
}
else if(c[i]==2)
{
if(e[i]+1>q[2])
{
cout << 70 << endl;
}
else
{
cout << 75 << endl;
}
}
else if(c[i]==1)
{
if(e[i]+1>q[1])
{
cout << 60 << endl;
}
else
{
cout << 65 << endl;
}
}
}
cout << endl;
}
return 0;
}