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.
思路：结构体存储数据，先根据分数排序，在每个分数段（除100）根据时间排序，时间在1/2之前的分数加5，再对原来的序号排序，输出。
输出格式：here is a blank line after each case.
附代码
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 struct Stu { 7 int num; 8 int sco; 9 int tim; 10 }; 11 12 int cmp1(Stu s1,Stu s2) { 13 return s1.tim<s2.tim; 14 } 15 int cmp2(Stu s1,Stu s2) { 16 return s1.num<s2.num; 17 } 18 int cmp3(Stu s1,Stu s2) { 19 return s1.sco>s2.sco; 20 } 21 int main() { 22 23 freopen("C:\CODE\in.txt", "r", stdin); 24 //freopen("C:\CODE\out.txt","w",stdout); 25 int n,d,m,s; 26 bool flg=false; 27 char ch[100]; 28 29 while(~scanf("%d",&n)&&n>=1) { 30 getchar(); 31 int num[10]= {0}; 32 struct Stu stu1[105]; 33 for(int i=0; i<n; i++) { 34 gets(ch); 35 stu1[i].num=i+1; 36 stu1[i].sco=(ch[0]-'0')*10+50; 37 d=(ch[2]-'0')*10+(ch[3]-'0'); 38 m=(ch[5]-'0')*10+(ch[6]-'0'); 39 s=(ch[8]-'0')*10+(ch[9]-'0'); 40 stu1[i].tim=d*3600+m*60+s; 41 num[ch[0]-'0']++; 42 } 43 sort(stu1,stu1+n,cmp3);//分数升序 44 int p=num[5]; 45 for(int i=4; i>=1; i--) { 46 if(num[i]>=3) 47 sort(stu1+p,stu1+p+num[i],cmp1); 48 for(int j=0; j<num[i]/2; j++) { 49 stu1[p+j].sco+=5; 50 } 51 p+=num[i]; 52 } 53 54 sort(stu1,stu1+n,cmp2); 55 56 flg=true; 57 for(int i=0; i<n; i++) { 58 printf("%d ",stu1[i].sco); 59 stu1[i].num=0; 60 stu1[i].sco=0; 61 stu1[i].tim=0; 62 } 63 printf(" "); 64 } 65 66 fclose(stdin); 67 return 0; 68 }