1012. The Best Rank (25)
2016-02-18 08:51
357 查看
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 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
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
999999
Sample Output
1 C
1 M
1 E
1 A
3 A
N/A
使用了二分法查找成绩。因为中间有一次排序,本来以为会超时,但是A过了,应该是时间复杂度要求不高。
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
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
999999
Sample Output
1 C
1 M
1 E
1 A
3 A
N/A
使用了二分法查找成绩。因为中间有一次排序,本来以为会超时,但是A过了,应该是时间复杂度要求不高。
#include<iostream> #include<vector> #include<string> #include<algorithm> using namespace std; vector<int> c[4]; int n,m; struct node{ int c,m,e,a; string name; }; bool cmp(int a,int b){ return a>b; } int find(int t,int k){ int l=0,r=n-1,mid; while(l!=r){ mid=(l+r)/2; if(t<c[k][mid]) l=mid+1; else r=mid; } while(r>0&&c[k][r-1]==t){ r--; } return r+1; } int main(){ vector<node> student; cin>>n>>m; for(int i=0;i<n;i++) { int t,ta; node st; cin>>st.name; for(int j=0;j<3;j++) { cin>>t; c[j].push_back(t); } st.c=c[0][c[0].size()-1]; st.m=c[1][c[1].size()-1]; st.e=c[2][c[2].size()-1]; st.a=(st.m+st.e+st.c)/3; student.push_back(st); c[3].push_back(st.a); } for(int i=0;i<=3;i++) { sort(c[i].begin(),c[i].end(),cmp); } while(m--){ string name; cin>>name; int sc,sm,se,sa,i; for( i=0;i<student.size();i++) { if(name==student[i].name){ sc=find(student[i].c,0); sm=find(student[i].m,1); se=find(student[i].e,2); sa=find(student[i].a,3); if(sa<=min(se,sm)&&sa<=sc) { printf("%d A\n",sa);break; } else if(sc<=min(sa,se)) { printf("%d C\n",sc);break; } else if(sm<=se){ printf("%d M\n",sm);break; } else { printf("%d E\n",se);break; } } } if(i==student.size()) printf("N/A\n"); } return 0; }
相关文章推荐
- 【5-3】hadoop集群搭建
- 拼接字符串改为 @字段名,防止注入攻击
- EF-DEMO
- 程序在后台执行还有对应的tty么
- js金额输入规则
- 【HDU 5578】Friendship of Frog
- hadoop能用到的系统端口
- shell脚本实现从某一行开始附加内容
- 6410 qtopia2.2.0开发:二 编译ARM qtopia2.2.0
- LeeCode_136 Single Number
- svn的分支和合并相关资料
- iscroll.js的上拉下拉刷新时无法回弹的解决方法
- java中字符串转换为日期并且格式化
- 软件开发技术:流畅表达(fluent style)
- PAT1046
- 转载:oracle用户创建及权限设置
- 4检测平台
- Ke模拟器kemulator 1.0 绿色中文版
- CLRS 14.1动态顺序统计
- 网站常见的反爬虫和应对方法 + [评论]