hdu 1084 排序

#include <stdio.h>

#include <iostream>

#include <string>

#include <stdlib.h>

using namespace std;

#define mark student

#define que solve

struct	student

{

int solve;

int time;

int grade;

int set;

};

int cmp(const void *a, const void *b )

{

if((*(student *)a).solve != (*(student *)b).solve)

return (*(mark *)b).que-(*(mark *)a).que;

return (*(mark *)a).time-(*(mark *)b).time;

}

int main()

{

int n;

while( ~scanf("%d", &n)	&& n!= -1)

{

student list[110];

student *p[110];

int num[6];

memset(num, 0, sizeof(num) );

for(int i=0; i<n; i++)

	{

int hour, minute, second;

scanf("%d %d:%d:%d", &list[i].solve, &hour, &minute, &second);

list[i].time = hour*60*60 +minute * 60 +second;

num[list[i].solve] ++;

list[i].set =i;

}

qsort(list, n, sizeof(list[0]), cmp);

int count =0;

for(int i=5; i>=0; i--)

	{

if(i == 0)

for(int i=0; i<num[0]; i++)		list[count++].grade = 50;

else if(i == 5)

for(int i=0; i<num[5]; i++)		list[count++].grade = 100;

else

{

for(int j=1; j<=num[i]; j++)

	{

if(j<=num[i]/2)

list[count++].grade = (i+5)*10+5;

		else

list[count++].grade = (i+5)*10;

}

}

}

for(int i=0; i<n; i++)

	{

for(int j=0; j<n; j++)

{

if(list[j].set == i)	printf("%d\n", list[j].grade);

}

}

printf("\n");

}

return 0;

}

/*

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!

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

*/

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