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) onlywhen 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! InputInput 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. OutputOutput 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

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&
bf11
&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;}