HDOJ 1084 What Is Your Grade?

What Is Your Grade?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10630    Accepted Submission(s): 3299

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

45 06:30:174 07:31:274 08:12:124 05:23:1315 06:30:17-1

 

Sample Output

100909095100

 

Author

lcy

 

……一开始看错了一句很重要的话（做了同样多题目得人数量半数以上能的 X5 分，这句话），WA了几次。

不是很难，看代码应该就能懂，所以就不写题解了。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct Stu{int ans,g,id;char ti[25];}a[111];bool cmp1(Stu a,Stu b){if(a.ans!=b.ans)return a.ans>b.ans;return strcmp(a.ti,b.ti)<0;}bool cmp2(Stu a,Stu b){return a.id<b.id;}int main(){int n,i;while(scanf("%d",&n)&&n!=-1){int s1,s2,s3,s4;s1=s2=s3=s4=0;for(i=0;i<n;i++){scanf("%d%s",&a[i].ans,a[i].ti);a[i].id=i;if(a[i].ans==4) s4++;if(a[i].ans==3) s3++;if(a[i].ans==2) s2++;if(a[i].ans==1) s1++;}sort(a,a+n,cmp1);int p,q,r,s;p=q=r=s=0;for(i=0;i<n;i++){if(a[i].ans==5)a[i].g=100;if(a[i].ans==4){if(p<s4/2){a[i].g=95;p++;}elsea[i].g=90;}if(a[i].ans==3){if(q<s3/2){a[i].g=85;q++;}elsea[i].g=80;}if(a[i].ans==2){if(r<s2/2){a[i].g=75;r++;}elsea[i].g=70;}if(a[i].ans==1){if(s<s1/2){a[i].g=65;s++;}elsea[i].g=60;}if(a[i].ans==0)a[i].g=50;}sort(a,a+n,cmp2);for(i=0;i<n;i++)printf("%d\n",a[i].g);printf("\n");}return 0;}