http://ac.jobdu.com/problem.php?pid=1014
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- 题目描述:
- 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。
- 输入:
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。
- 输出:
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。
- 样例输入:
4 5 2510 10 12 13 15CS004 3 5 1 3CS003 5 2 4 1 3 5CS002 2 1 2CS001 3 2 3 51 2 4010 30CS001 1 22 3 2010 10 10CS000000000000000001 0CS000000000000000002 2 1 20
- 样例输出:
3CS003 60CS001 37CS004 3701CS000000000000000002 20
- #include<iostream>
- #include<cstdio>
- #include<memory.h>
- #include<algorithm>
- #include<cstring>
- #include<cstdlib>
- using namespace std;
- struct Node{
- char s[21];
- int grade;
- bool operator < (const Node & node)const{
- if(grade!=node.grade)
- return grade > node.grade;
- return strcmp(s, node.s) < 0;
- }
- };
- Node nodes[1001];
- int grade[11];
- int main(){
- //freopen("in.txt", "r", stdin);
- int n, m, g, k, t, sum;
- while(cin>>n, n){
- cin>>m>>g;
- for(int i=1;i<=m;++i){
- scanf("%d", grade+i);
- }
- for(int i=0;i<n;++i){
- cin>>nodes[i].s;
- cin>>k;
- sum = 0;
- for(int j=0;j<k;++j){
- scanf("%d", &t);
- sum += grade[t];
- }
- nodes[i].grade = sum;
- }
- sort(nodes, nodes+n);
- int i;
- for(i=0;i<n;++i){
- if(nodes[i].grade < g){
- break;
- }
- }
- printf("%d\n", i);
- for(int j=0;j<i;++j){
- printf("%s %d\n", nodes[j].s, nodes[j].grade);
- }
- }
- //fclose(stdin);
- return 0;
- }
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