浙大机试2005:排名

来源:互联网 发布:六小龄童 孙悟空 知乎 编辑:程序博客网 时间:2024/04/29 12:06
Problem Description
今天的上机考试虽然有实时的Ranklist,但上面的排名只是根据完成的题数排序,没有考虑 
每题的分值,所以并不是最后的排名。给定录取分数线,请你写程序找出最后通过分数线的 
考生,并将他们的成绩按降序打印。 


Input
测试输入包含若干场考试的信息。每场考试信息的第1行给出考生人数N ( 0 < N 
< 1000 )、考题数M ( 0 < M < = 10 )、分数线(正整数)G;第2行排序给出第1题至第M题的正整数分值;以下N行,每行给出一 
名考生的准考证号(长度不超过20的字符串)、该生解决的题目总数m、以及这m道题的题号 
(题目号由1到M)。 
当读入的考生人数为0时,输入结束,该场考试不予处理。 


Output
对每场考试,首先在第1行输出不低于分数线的考生人数n,随后n行按分数从高 
到低输出上线考生的考号与分数,其间用1空格分隔。若有多名考生分数相同,则按他们考 
号的升序输出。 


Sample Input
4 5 25
10 10 12 13 15
CS004 3 5 1 3
CS003 5 2 4 1 3 5
CS002 2 1 2
CS001 3 2 3 5
1 2 40
10 30
CS001 1 2
2 3 20
10 10 10
CS000000000000000001 0
CS000000000000000002 2 1 2
0
 
Sample Output
3
CS003 60
CS001 37
CS004 37
0
1

CS000000000000000002 20


#include <stdio.h>#include <string.h>#define MAXN 1000#define MAXM 11typedef struct{char id[21]; //准考证号int sum;//总分}EXAMINEE;EXAMINEE examinee[MAXN];int score[MAXM];void init(int n){int i;for(i = 1;i <= n;++i){examinee[i].sum = 0;}}int main(){int N,M,G;int i,j,problem,m,cnt;int flag = 0;EXAMINEE tmp;while(scanf("%d%d%d",&N,&M,&G) && N){init(N);for(i = 1;i <= M;++i){scanf("%d",&score[i]);}for(i = 1;i <= N;++i){scanf("%s%d",&examinee[i].id,&m);for(j = 1;j <= m;++j){//输入题号时就把总分计算出来scanf("%d",&problem);examinee[i].sum = examinee[i].sum + score[problem];}}j = 1;for(i = 1;i <= N;++i){if(examinee[i].sum >= G)examinee[j++] = examinee[i];}cnt = j - 1;if(cnt == 0){printf("0\n");continue;}for(i = 1;i <= cnt;++i){for(j = 1;j < cnt - i + 1;++j){if( (examinee[j].sum < examinee[j+1].sum) ||((examinee[j].sum == examinee[j+1].sum)&&(strcmp(examinee[j].id, examinee[j+1].id)>0)) ){tmp = examinee[j];examinee[j] = examinee[j+1];examinee[j+1] = tmp;flag = 1;}}if(!flag) break;}printf("%d\n",cnt);for(i = 1;i <= cnt;++i){printf("%s %d\n",examinee[i].id,examinee[i].sum);}}return 0;}





0 0
原创粉丝点击