http://acm.hdu.edu.cn/showproblem.php?pid=1084
What Is Your Grade?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9580 Accepted Submission(s): 2940
Problem 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!
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.
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
不想多解释，水题一道，题没看错就行，这么一道水题我竟wa了很多次，很是忧桑。。。
直接附上代码
#include <stdio.h>
#include <stdlib.h>
#include <string.h> #define N 110 using namespace std; struct st
{
int id, x, h, y;
} node[N]; int cmp(const void *a, const void *b)
{
st *s1 = (st *)a, *s2 = (st *)b;
if(s1->x == s2->x)
return s1->h - s2->h;
else
return s2->x - s1->x;
} int cmp1(const void *a, const void *b)
{
return *(int *)a - *(int *)b;
} int main()
{
int i, a1, a2, a3, a4, n, h, m, s;
while(scanf("%d", &n), n != -)
{
memset(node, , sizeof(node));
a1 = a2 = a3 = a4 = ;
for(i = ; i < n ; i++)
{
scanf("%d", &node[i].x);
scanf("%d:%d:%d", &h, &m, &s);
node[i].h = h * + m * + s;
node[i].id = i;
if(node[i].x == )
a1++;
else if(node[i].x == )
a2++;
else if(node[i].x == )
a3++;
else if(node[i].x == )
a4++;
}
qsort(node, n, sizeof(node[]), cmp);
a1 /= , a2 /= , a3 /= , a4 /= ;
for(i = ; i < n ; i++)
{
if(node[i].x == )
node[i].y = ;
else if(node[i].x == )
{
if(a4 != )
{
node[i].y = ;
a4--;
}
else
node[i].y = ;
}
else if(node[i].x == )
{
if(a3 != )
{
node[i].y = ;
a3--;
}
else
node[i].y = ;
}
else if(node[i].x == )
{
if(a2 != )
{
node[i].y = ;
a2--;
}
else
node[i].y = ;
}
else if(node[i].x == )
{
if(a1 != )
{
node[i].y = ;
a1--;
}
else
node[i].y = ;
}
else
node[i].y = ;
}
qsort(node, n, sizeof(node[]), cmp1);
for(i = ; i < n ; i++)
printf("%d\n", node[i].y);
printf("\n"); }
return ;
}