首页 > ACM题库 > 九度OJ > 九度-1324-The Best Rank[基础排序题]
2014
01-16

九度-1324-The Best Rank[基础排序题]

题目来自九度:2011年九度互动社区IT名企招聘上机考试热身赛

题目1324:The Best Rank

题目描述:
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.

输入:
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.
输出:
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”.

样例输入:
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
样例输出:
1 C
1 M
1 E
1 A
3 A
N/A

基础排序题:假设四个人的C成绩为 90 90 80 70 则排名为 1 1 3 4

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
struct stu {
    stu(string& n,int i,int j,int k):id(n) {
        c[0]=(i+j+k)/3;
        c[1]=i;c[2]=j;c[3]=k;
        }
    string id;
    int c[4],bst,tp,r[4];
    };
bool p1(const stu a,const stu b) {
    if(a.c[0]>b.c[0])return true;
    else return false;
    }
bool p2(const stu a,const stu b) {
    if(a.c[1]>b.c[1])return true;
    else return false;
    }
bool p3(const stu a,const stu b) {
    if(a.c[2]>b.c[2])return true;
    else return false;
    }
bool p4(const stu a,const stu b) {
    if(a.c[3]>b.c[3])return true;
    else return false;
    }
int main() {
    int n,m,c,e,im;
    string id;
    bool (*cp[4])(const stu,const stu);
    cp[0]=p1;cp[1]=p2;cp[2]=p3;cp[3]=p4;
    while(cin>>n>>im) {
        vector < stu > vs;
        while(n--) {
            cin>>id>>c>>m>>e;
            vs.push_back(stu(id,c,m,e));
            }
        for(int fc=0; fc<4; ++fc) {

            sort(vs.begin(),vs.end(),cp[fc]);
            vs[0].r[fc]=1;
            for(int i=1; i<vs.size(); ++i)
                if(vs[i].c[fc]==vs[i-1].c[fc])
                    vs[i].r[fc]=vs[i-1].r[fc];
                else vs[i].r[fc]=i+1;
            }
        for(int i=0;i<vs.size();++i){
                vs[i].bst=vs[i].r[0];vs[i].tp=0;
                for(int k=1;k<4;++k)
                    if(vs[i].r[k]<vs[i].bst){
                        vs[i].bst=vs[i].r[k];
                        vs[i].tp=k;
                        }
            }
        while(im--) {
            bool ok=false;
            cin>>id;
            for(int i=0; i<vs.size(); ++i) {
                if(vs[i].id==id) {
                    ok=true;
                    cout<<vs[i].bst<<' ';
                    switch(vs[i].tp) {
                        case 0:cout<<"A"<<endl;break;
                        case 1:cout<<"C"<<endl;break;
                        case 2:cout<<"M"<<endl;break;
                        case 3:cout<<"E"<<endl;break;
                        }
                    break;
                    }
                }
            if(!ok)cout<<"N/A"<<endl;
            }
        }
    }

 


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  2. 问题3是不是应该为1/4 .因为截取的三段,无论是否能组成三角形, x, y-x ,1-y,都应大于0,所以 x<y,基础应该是一个大三角形。小三角是大三角的 1/4.