pta--1012 The Best Rank (25)（结构体&&cmp函数重写）
To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematics (Calculus or Linear Algebra), and E - English. At the mean time, we encourage students by emphasizing on their best ranks -- that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.
For example, The grades of C, M, E and A - Average of 4 students are given as the following:
StudentID C M E A
310101 98 85 88 90
310102 70 95 88 84
310103 82 87 94 88
310104 91 91 91 91Then the best ranks for all the students are No.1 since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.
Input
Each input file contains one test case. Each case starts with a line containing 2 numbers N and M (<=2000), which are the total number of students, and the number of students who would check their ranks, respectively. Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of [0, 100]) of that student in the order of C, M and E. Then there are M lines, each containing a student ID.
Output
For each of the M students, print in one line the best rank for him/her, and the symbol of the corresponding rank, separated by a space.
The priorities of the ranking methods are ordered as A > C > M > E. Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.
If a student is not on the grading list, simply output "N/A".
Sample Input
5 6
310101 98 85 88
310102 70 95 88
310103 82 87 94
310104 91 91 91
310105 85 90 90
310101
310102
310103
310104
310105
999999Sample Output
1 C
1 M
1 E
1 A
3 A
N/A【分析】
1. 要注意如果有排名相同的情况。如1、1、3、4而不是1、2、3、4；这个点我改了好久；
2. 我重写了4个cmp函数，在调用以后进行排名的时候，新定义一个hh来记录重复值的次数，注意每次sort前要对hh重赋值为1，不然会出错的。
3. 主要还是细节的处理吧。慢慢来，，，佛性思想，急不得呀hhhh
ppppps///代码有点长有点繁琐
#include
using namespace std;
struct stu{
int id;
int c,m,e;
int crank,mrank,erank,arank;
double a;
}st[2000];
bool cmp1(stu x,stu y)
{
return x.a>y.a;
}
bool cmp2(stu x,stu y)
{
return x.c>y.c;
}
bool cmp3(stu x,stu y)
{
return x.m>y.m;
}
bool cmp4(stu x,stu y)
{
return x.e>y.e;
}
int ff(int a,int b,int c,int d)
{
int aa[4]={a,b,c,d};
sort(aa,aa+4);
if(a==aa[0])return 1;
if(b==aa[0])return 2;
if(c==aa[0])return 3;
if(d==aa[0])return 4;
// a=aa[3],b=aa[2],c=aa[1],d=aa[0];
return -1;
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
int hh=0;
for(int i=0;i