HPUOJ---2017寒假作业--专题0/H-What Is Your Grade（switch语句）

H - What Is Your Grade?

 

“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

思路：题目说的很清楚，0道50分，5道100分，1道至4道，根据提交时间 ，同样的题数，前半部分比后半部分

多五分。 

代码一：自己的思路（很烂很菜的）
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct stu{int n;char time[10];int turn;int k;int p;  } data[100+25]; bool cmp1(stu A,stu B) { if(A.n==B.n) { return strcmp(A.time,B.time)<0;}                     //字符串排序就应该这样，一定要记清楚了  return A.n>B.n; } bool cmp2(stu A,stu B) { return A.turn<B.turn; } int main() { int N,i,k[100+25],a,b,c,d,j,m; while(scanf("%d",&N),N>=0) { a=1,b=1,c=1,d=1,j=-1; for(i=0;i<N;i++) { scanf("%d %s",&data[i].n,data[i].time); data[i].turn=i;}sort(data,data+N,cmp1);    for(i=0;i<N;i++){ if(data[i].n==5) {     data[i].k=100;     j=i;    } else if(data[i].n==data[i+1].n&&data[i].n==4) a++; else if(data[i].n==data[i+1].n&&data[i].n==3)     b++; else if(data[i].n==data[i+1].n&&data[i].n==2)     c++; else if(data[i].n==data[i+1].n&&data[i].n==1)     d++; else     data[i].k=50;}m=j+1;for(i=m;i<m+a/2;i++)     data[i].k=95;for(i=m+a/2;i<m+a;i++) data[i].k=90;for(i=m+a;i<m+a+b/2;i++) data[i].k=85;for(i=m+a+b/2;i<m+a+b;i++)    data[i].k=80;for(i=m+a+b;i<m+a+b+c/2;i++)     data[i].k=75;for(i=m+a+b+c/2;i<m+a+b+c;i++)    data[i].k=70;for(i=m+a+b+c;i<m+a+b+c+d/2;i++)    data[i].k=65;for(i=m+a+b+c+d/2;i<m+a+b+c+d;i++) data[i].k=60;for(i=0;i<N;i++)    data[i].p=data[i].k; sort(data,data+N,cmp2);//因为要按原顺序输出，所以把已经被打乱的顺序重新再排 for(i=0;i<N;i++)printf("%d\n",data[i].p);printf("\n"); } return 0; }
代码二：借鉴某大神的思路。switch语句，很厉害。

#include<cstdio>#include<cstring> #include<algorithm>using namespace std;struct stu{int n;char time[10];int turn;int k;}data[100+12];bool cmp1(stu A,stu B){if(A.n==B.n)   return strcmp(A.time,B.time)<0;return A.n>B.n; }bool cmp2(stu A,stu B){return A.turn<B.turn;                              //最后要按原序输出 ，将顺序排列，最后回归原序}int main(){int N,i,p[100+10],a;                       //用a来代表同样题数提交的顺序（厉害）while(scanf("%d",&N),N>=0){a=0;memset(p,0,sizeof(p));for(i=0;i<N;i++){scanf("%d %s",&data[i].n,data[i].time);data[i].turn=i;p[data[i].n]++;//用这个方法来计算统计做对某一题数的人数，更加方便，不易想到 //}for(i=0;i<N;i++){switch(data[i].n){case 0:data[i].k=50;break;case 1:data[i].k=60;break;case 2:data[i].k=70;break;case 3:data[i].k=80;break;case 4:data[i].k=90;break;case 5:data[i].k=100;break; }if(data[i].n>0&&data[i].n<5){if(p[data[i].n]<=1||a<p[data[i].n]/2)                  //这一点很不容易想到啊 //{data[i].k+=5;a++;                             //前几个的分数较高，那咋办呢，不用你再那样都算开，直接引入一个新的变量就可以了，// }if(data[i].n!=data[i+1].n)                     //和下面的不一样了就要归0了，进入下一个 //    a=0;}}sort(data,data+N,cmp2);for(i=1;i<N;i++)   printf("%d\n",data[i].k);printf("\n");}return 0;                            }