HDU 1084 What Is Your Grade?

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct Stu {
    int num;
    int sco;
    int tim;
};

int cmp1(Stu s1,Stu s2) {
    return s1.tim<s2.tim;
}
int cmp2(Stu s1,Stu s2) {
    return s1.num<s2.num;
}
int cmp3(Stu s1,Stu s2) {
    return s1.sco>s2.sco;
}
int main() {

    freopen("C:\\CODE\\in.txt", "r", stdin);
    //freopen("C:\\CODE\\out.txt","w",stdout);
    int n,d,m,s;
    bool flg=false;
    char ch[100];

    while(~scanf("%d",&n)&&n>=1) {
        getchar();
        int num[10]= {0};
        struct Stu stu1[105];
        for(int i=0; i<n; i++) {
            gets(ch);
            stu1[i].num=i+1;
            stu1[i].sco=(ch[0]-'0')*10+50;
            d=(ch[2]-'0')*10+(ch[3]-'0');
            m=(ch[5]-'0')*10+(ch[6]-'0');
            s=(ch[8]-'0')*10+(ch[9]-'0');
            stu1[i].tim=d*3600+m*60+s;
            num[ch[0]-'0']++;
        }
        sort(stu1,stu1+n,cmp3);//分数升序
        int p=num[5];
        for(int i=4; i>=1; i--) {
            if(num[i]>=3)
                sort(stu1+p,stu1+p+num[i],cmp1);
            for(int j=0; j<num[i]/2; j++) {
                stu1[p+j].sco+=5;
            }
            p+=num[i];
        }

        sort(stu1,stu1+n,cmp2);

        flg=true;
        for(int i=0; i<n; i++) {
            printf("%d\n",stu1[i].sco);
            stu1[i].num=0;
            stu1[i].sco=0;
            stu1[i].tim=0;
        }
        printf("\n");
    }

    fclose(stdin);
    return 0;
}