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  • PAT 甲级 1036 Boys vs Girls(20)

    https://pintia.cn/problem-sets/994805342720868352/problems/994805453203030016

    This time you are asked to tell the difference between the lowest grade of all the male students and the highest grade of all the female students.

    Input Specification:

    Each input file contains one test case. Each case contains a positive integer N, followed by N lines of student information. Each line contains a student's namegenderID and grade, separated by a space, where name and ID are strings of no more than 10 characters with no space, gender is either F (female) or M (male), and grade is an integer between 0 and 100. It is guaranteed that all the grades are distinct.

    Output Specification:

    For each test case, output in 3 lines. The first line gives the name and ID of the female student with the highest grade, and the second line gives that of the male student with the lowest grade. The third line gives the difference gradeF​​gradeM​​. If one such kind of student is missing, output Absent in the corresponding line, and output NA in the third line instead.

    Sample Input 1:

    3
    Joe M Math990112 89
    Mike M CS991301 100
    Mary F EE990830 95
    

    Sample Output 1:

    Mary EE990830
    Joe Math990112
    6
    

    Sample Input 2:

    1
    Jean M AA980920 60
    

    Sample Output 2:

    Absent
    Jean AA980920
    NA


    时间复杂度:$O(N)$

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    int N;
    
    struct Students {
        char name[20];
        char sex[5];
        char num[20];
        int Grade;
    }students[100100];
    
    /*bool cmp(const Students& a, const Students& b) {
        if(a.sex != b.sex)
            return a.sex > b.sex;
        else
            return a.Grade > b.Grade;
    }*/
    
    int main() {
        scanf("%d", &N);
        for(int i = 1; i <= N; i ++)
            scanf("%s%s%s%d", students[i].name, students[i].sex, students[i].num, &students[i].Grade);
    
    
        int m = 0, f = 0;
        //sort(students + 1, students + 1 + N, cmp);
    
        int s1 = 999, s2 = -999;
        int temp1 = 0, temp2 = 0;
        for(int i = 1; i <= N; i ++) {
            if(strcmp(students[i].sex, "M") == 0) {
                if(students[i].Grade < s1) {
                    s1 = students[i].Grade;
                    temp1 = i;
                }
            } else {
                if(students[i].Grade > s2) {
                    s2 = students[i].Grade;
                    temp2 = i;
                }
            }
        }
    
        for(int i = 1; i <= N; i ++) {
            if(strcmp(students[i].sex, "M") == 0)
                m ++;
            else
                f ++;
        }
    
        if(f == 0)
            printf("Absent
    %s %s
    NA
    ", students[temp1].name, students[temp1].num);
        else if(m == 0)
            printf("%s %s
    Absent
    NA
    ", students[temp2].name, students[temp2].num);
        else {
            printf("%s %s
    ", students[temp2].name, students[temp2].num);
            printf("%s %s
    ", students[temp1].name, students[temp1].num);
            printf("%d
    ", students[temp2].Grade - students[temp1].Grade);
        }
    
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9649082.html
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