题目
1012 The Best Rank （25 分）
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 Algrbra), 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 91
Then 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 Specification:
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 Specification:
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
999999
Sample Output:
1 C
1 M
1 E
1 A
3 A
N/A
思路
题目要求是给出每个人的最好排名，所以总体思路是排序，查找。
时限不严，排序用冒泡，查找用遍历就可以通过。
但题中有些地方值得注意：
题中对并列有要求，则查找可以按照优先级顺序进行查找，则在找最小次序时，不需要再考虑优先级。
记录要查找的人的最好排名的数组需要特殊初始化。不能默认为0，会导致查找出错。需要初始化为N，不是M。
测试点2考虑了多个人并列的问题，则需要记录每个成绩对应的排名，不能仅仅是通过下标来标识。
#include <stdio.h>
#include <string.h>typedef struct G {int grade;char id[7];int order;
}grade;void sort(grade a[],int len){grade tem ;int i , j ;for(i=0;i<len;i++){for(j=0;j<len-i-1;j++){if(a[j+1].grade>a[j].grade){strcpy(tem.id , a[j+1].id);tem.grade = a[j+1].grade;strcpy( a[j+1].id , a[j].id);a[j+1].grade = a[j].grade;strcpy(a[j].id , tem.id);a[j].grade = tem.grade;a[j].order = j;a[j+1].order = j+1 ;}else if(a[j+1].grade== a[j].grade){a[j+1].order = a[j].order ;}else{a[j].order = j;a[j+1].order = j+1 ; }}}
}int main(){int i , j ,k,n ,m ,flag;if( 2 != scanf("%d %d",&n,&m))return -2;grade g[4][n];char name[7];int r1[m];char r2[m][20];//cchar cn[] ={"ACME\0"};for(i=0;i<m;i++){r1[i] = n; // !!!!!not m}for(i=0;i<n;i++){if(4 != scanf("%s %d %d %d",name,&(g[1][i].grade),&(g[2][i].grade),&(g[3][i].grade)))return -2;strcpy(g[0][i].id , name);strcpy(g[1][i].id , name);strcpy(g[2][i].id , name);strcpy(g[3][i].id , name);g[0][i].grade = g[3][i].grade+g[1][i].grade+g[2][i].grade;}sort(g[0],n);sort(g[1],n);sort(g[2],n);sort(g[3],n);for(i=0;i<m;i++){flag = 0;if(1 != scanf("%s",name))return -2;for(j=0; j<4;j++){for(k=0;k<n;k++){if(strcmp(name,g[j][k].id)==0){flag = 1;if(g[j][k].order < r1[i]){r1[i] = g[j][k].order ;sprintf(r2[i],"%d %c",g[j][k].order +1,cn[j]);}}}if(!flag){strcpy(r2[i],"N/A\0");break;}}}printf("%s",r2[0]);for(i=1;i<m;i++){printf("\n%s",r2[i]);}
}