杭电OJ—— 1084 What Is Your Grade?
What Is Your Grade?
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
Author
lcy
题目不是很难，注意一下输出格式还有一些别的东西就可以了！我下面写的程序貌似长了一点，不过意思明了，凑合着看吧！很久没写了，练练手！
#include <iostream>
#include <algorithm>
using namespace std;
const int SIZE = 101;
typedef struct /*结构体，用于存储输入的数据*/
{
int solved; /*解决问题的数目*/
int located; /*学生的位置*/
int h, m, s; /*所花费的时间*/
int soc; /*最终的成绩*/
}time;
bool cmp(time &bi1, time &bi2) /*比较函数，用于排序*/
{
if (bi1.h > bi2.h) return false;
else if (bi1.h < bi2.h) return true;
else
{
if (bi1.m > bi2.m) return false;
else if (bi1.m < bi2.m) return true;
else
{
if (bi1.s > bi2.s) return false;
else if (bi1.s < bi2.s) return true;
}
}
}
int main()
{
int num;
time stu[SIZE];
time sortArr[4][SIZE];
int p, q , r, k, i;
while((cin >> num) && (num >= 0))
{
p = q = r = k = 0;
for (i = 0; i < num; i++)
{
cin >> stu[i].solved;
scanf ("%d:%d:%d", &stu[i].h, &stu[i].m, &stu[i].s);
stu[i].located = i; /*记录下输入时的位置*/
/*下面的思路就是：分数能处理的就处理，如解决了5道题（100）和0道题（50）
不能处理的话，我们先将该数分类存入一个数组，之后再做处理
*/
if (stu[i].solved == 5)
{
stu[i].soc = 100;
}
if (stu[i].solved == 4)
{
sortArr[0][p] = stu[i];
p++;
}
if (stu[i].solved == 3)
{
sortArr[1][q] = stu[i];
q++;
}
if (stu[i].solved == 2)
{
sortArr[2][r] = stu[i];
r++;
}
if (stu[i].solved == 1)
{
sortArr[3][k] = stu[i];
k++;
}
if (stu[i].solved == 0)
{
stu[i].soc = 50;
}
}
if (p >= 1)
{
sort (sortArr[0], sortArr[0] + p, cmp); /*先排序，按照耗时从小到大排序*/
for (i = 0; i < p / 2; i++) /*前半部分得高分*/
{
stu[sortArr[0][i].located].soc = 95;
}
for (; i < p; i++) /*后半部分得低分*/
{
stu[sortArr[0][i].located].soc = 90;
}
}
/*以下类似*/
if (q >= 1)
{
sort (sortArr[1], sortArr[1] + q, cmp);
for (i = 0; i < q / 2; i++)
{
stu[sortArr[1][i].located].soc = 85;
}
for (; i < q; i++)
{
stu[sortArr[1][i].located].soc = 80;
}
}
if (r >= 1)
{
sort (sortArr[2], sortArr[2] + r, cmp);
for (i = 0; i < r / 2; i++)
{
stu[sortArr[2][i].located].soc = 75;
}
for (; i < r; i++)
{
stu[sortArr[2][i].located].soc = 70;
}
}
if (k >= 1)
{
sort (sortArr[3], sortArr[3] + k, cmp);
for (i = 0; i < k / 2; i++)
{
stu[sortArr[3][i].located].soc = 65;
}
for (; i < k; i++)
{
stu[sortArr[3][i].located].soc = 60;
}
}
for (i = 0; i < num; i++)
{
cout << stu[i].soc << endl;
}
cout << endl;
}
system ("pause");
return 0;
}