What Is Your Grade?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9469 Accepted Submission(s): 2903
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
C语言程序代码
/*
题意::
给出5道题,若完成5道则给100分,若做4道题要分两种情况(1、按时间排名,在前半段
的给95分,后半部分给90分) 则以后以此类推有85,80,75,70,65,60分,当做0道题时
给50分。
解题思路::
这题本身是个简单的结构体,但难的是要达到分阶段给分这个目的是不容易的
此代码所用方式为定义四个变量用来分别记录做4,3,2,1道题的人数,然后
将作出相同的题数的人的时间从小到大排序,前半部分分别给95 ,85,75,65分
当然,还有一个重点是,要记录学生的位置,然后按相应位置给出对应分数。
*/
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct student
{
int qm;/*解决问题的数目*/
int lo;/*学生的位置*/
char sj[100];/*所花费的时间*/
int fs;/*最终的成绩*/
}stu[101];
bool c1(student x,student y)
{
return strcmp(x.sj,y.sj)<0;
}
int main(){
student s[4][101];
student s1[101];
int t,n,m,i,j,a,b,c,d;
while(scanf("%d",&t)!=EOF)
{
if(t==-1)
break;
a=b=c=d=0;
for(i=0;i<t;i++)
{
scanf("%d %s",&stu[i].qm,stu[i].sj);
stu[i].lo=i;
if(stu[i].qm==5)
{
stu[i].fs=100;
}
if(stu[i].qm==4)
{
s[0][a]=stu[i];
a++;
}
if(stu[i].qm==3)
{
s[1][b]=stu[i];
b++;
}
if(stu[i].qm==2)
{
s[2][c]=stu[i];
c++;
}
if(stu[i].qm==1)
{
s[3][d]=stu[i];
d++;
}
if(stu[i].qm==0)
{
stu[i].fs=50;
}
}
if(a>=1)
{
sort(s[0],s[0]+a,c1);
for(i=0;i<a/2;i++)
{
stu[s[0][i].lo].fs=95;
}
for(;i<a;i++)
{
stu[s[0][i].lo].fs=90;
}
}
if(b>=1)
{
sort(s[1],s[1]+b,c1);
for(i=0;i<b/2;i++)
{
stu[s[1][i].lo].fs=85;
}
for(;i<b;i++)
{
stu[s[1][i].lo].fs=80;
}
}
if(c>=1)
{
sort(s[2],s[2]+c,c1);
for(i=0;i<c/2;i++)
{
stu[s[2][i].lo].fs=75;
}
for(;i<c;i++)
{
stu[s[2][i].lo].fs=70;
}
}
if(d>=1)
{
sort(s[3],s[3]+d,c1);
for(i=0;i<d/2;i++)
{
stu[s[3][i].lo].fs=65;
}
for(;i<d;i++)
{
stu[s[3][i].lo].fs=60;
}
}
for(i=0;i<t;i++)
{
printf("%d\n",stu[i].fs);
}
printf("\n");
}
return 0;
}