• pat1036. Boys vs Girls (25)


    1036. Boys vs Girls (25)

    时间限制
    400 ms
    内存限制
    65536 kB
    代码长度限制
    16000 B
    判题程序
    Standard
    作者
    CHEN, Yue

    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 name, gender, ID 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
    

    提交代码

     1 #include<cstdio>
     2 #include<algorithm>
     3 #include<iostream>
     4 #include<cstring>
     5 #include<queue>
     6 #include<vector>
     7 #include<cmath>
     8 using namespace std;  
     9 struct node{
    10     char name[15],id[15];
    11     int grade;
    12     bool m;
    13     node(){
    14         m=false;
    15     }
    16 };
    17 bool cmp(node a,node b){
    18     return a.grade>b.grade;
    19 }
    20 int main(){
    21     //freopen("D:\input.txt","r",stdin);
    22     int n;
    23     scanf("%d",&n);
    24     node *stu=new node[n];
    25     int i,g=-1,b=-1;
    26     char t;
    27     for(i=0;i<n;i++){
    28         scanf("%s",stu[i].name);
    29         cin>>t;
    30         if(t=='M'){
    31             stu[i].m=true;
    32         }
    33         scanf("%s %d",stu[i].id,&stu[i].grade);
    34     }
    35     sort(stu,stu+n,cmp);
    36     for(i=0;i<n;i++){
    37         if(!stu[i].m){
    38             g=i;
    39             break;
    40         }
    41     }
    42     for(i=n-1;i>=0;i--){
    43         if(stu[i].m){
    44             b=i;
    45             break;
    46         }
    47     }
    48     if(g!=-1){
    49         printf("%s %s
    ",stu[g].name,stu[g].id);
    50     }
    51     else{
    52         printf("Absent
    ");
    53     }
    54     if(b!=-1){
    55         printf("%s %s
    ",stu[b].name,stu[b].id);
    56     }
    57     else{
    58         printf("Absent
    ");
    59     }
    60     if(g!=-1&&b!=-1){
    61         printf("%d
    ",stu[g].grade-stu[b].grade);
    62     }
    63     else{
    64         printf("NA
    ");
    65     }
    66     return 0;
    67 }
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  • 原文地址:https://www.cnblogs.com/Deribs4/p/4768111.html
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