HDU 1084:What Is Your Grade?
2018-01-31 19:26
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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
题意:输入n,表示n个人,第一个数表示做出了来的题数,第二个表示时间,做题数等于5得100分,4题时当做题时间排在前一半的时候得95,后一半得90,以此类推,题数等于0得0分。
/*放寒假以来第一次在没有空调的桌子旁写题,真的冷QAQ冷的没思路,看了大佬的题解才有了思路,代码也是模仿的
点击查看参(chao)考(xi)的代码QAQ */
我的代码
“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
题意:输入n,表示n个人,第一个数表示做出了来的题数,第二个表示时间,做题数等于5得100分,4题时当做题时间排在前一半的时候得95,后一半得90,以此类推,题数等于0得0分。
/*放寒假以来第一次在没有空调的桌子旁写题,真的冷QAQ冷的没思路,看了大佬的题解才有了思路,代码也是模仿的
点击查看参(chao)考(xi)的代码QAQ */
我的代码
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; struct wzy{ int point,I,NO; int h,m,s; }p[200]; int cmp(wzy u,wzy v) { if(u.point==v.point) { if(u.h==v.h){ if(u.m==v.m){ return u.s<v.s; } else return u.m<v.m; } else return u.h<v.h; } return u.point>v.point; } int cmp1(wzy u,wzy v) { return u.I<v.I; } int main() { int n,i; int k1,k2,k3,k4; int b1,b2,b3,b4; while(~scanf("%d",& 4000 ;n)) { k1=k2=k3=k4=1; b4=b3=b2=b1=0; if(n<=0) break; for(i=0;i<n;i++) { scanf("%d %d:%d:%d",&p[i].point,&p[i].h,&p[i].m,&p[i].s); p[i].I=i; if(p[i].point==4) b4++; if(p[i].point==3) b3++; if(p[i].point==2) b2++; if(p[i].point==1) b1++; } sort(p,p+n,cmp); for(i=0;i<n;i++)//记录每个题数所在的位置 (排名) { if(p[i].point==4) p[i].NO=k4++; if(p[i].point==3) p[i].NO=k3++; if(p[i].point==2) p[i].NO=k2++; if(p[i].point==1) p[i].NO=k1++; } sort(p,p+n,cmp1); for(i=0;i<n;i++) { if(p[i].point==5) printf("100\n"); if(p[i].point==4) { if(p[i].NO<=b4/2) printf("95\n"); else printf("90\n"); } if(p[i].point==3) { if(p[i].NO<=b3/2) printf("85\n"); else printf("80\n"); } if(p[i].point==2) { if(p[i].NO<=b2/2) printf("75\n"); else printf("70\n"); } if(p[i].point==1) { if(p[i].NO<=b1/2) printf("65\n"); else printf("60\n"); } if(p[i].point==0) printf("50\n"); } printf("\n"); } return 0; }
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