zoukankan      html  css  js  c++  java
  • hdu 4176

    Class Statistics

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 233    Accepted Submission(s): 153


    Problem Description
    The new principal of Woop Woop Public plans to meet the teaching team to discuss the performance of the classes/teachers and, being a bean counting fundamentalist, he wants to arm himself with some statistics for the meetings.
    Your task is to write a program that reads the pupils' marks in each class and generates performance reports for the principal prior to the meetings.
     
    Input
    The input starts with an integer K (1 <= K <= 100) indicating the number of classes on a line by itself. Each of the following K lines gives a class's data, which starts with an integer N (2 <= N <= 50) indicating the number of pupils in the class. The number of pupils is followed by their marks, given as integers, in the range of zero to one hundred, separated by single spaces.
     
    Output
    The report for each class consists of two lines.
    The first line consists of the sentence: "Class X", where X indicates the class number starting with the value of one.
    The second line reports the maximum class mark, minimum class mark and the largest difference between consecutive marks (when sorted in non-decreasing order) in the class using the formats shown in the sample below.
     
    Sample Input
    2 5 30 25 76 23 78 6 25 50 70 99 70 90
     
    Sample Output
    Class 1 Max 78, Min 23, Largest gap 46 Class 2 Max 99, Min 25, Largest gap 25
     
    Source
     
    Recommend
    lcy   |   We have carefully selected several similar problems for you:  4178 4177 4179 4180 4181
     
    #include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int a[500];
    int main(){
        int t;
        scanf("%d",&t);
        int cnt=0;
        while(t--){
            int n;
            cnt++;
            memset(a,0,sizeof(a));
            scanf("%d",&n);
            int temp=-1,temp1;
            for(int i=0;i<n;i++){
                scanf("%d",&a[i]);
    
    
                }
            sort(a,a+n);
            for(int i=1;i<n;i++){
    
                     temp1=a[i]-a[i-1];
                if(temp1>temp)
                    temp=temp1;
            }
            printf("Class %d
    ",cnt);
            printf("Max %d, Min %d, Largest gap %d
    ",a[n-1],a[0],temp);
    
        }
    
        return 0;
    }
  • 相关阅读:
    javaBean为什么要实现Serializable接口?
    OpenLayers3的WMS空间查询实现多个图层
    (WPS) 网络地理信息处理服务
    window.open跳过浏览器拦截
    linux改权限
    element-ui MessageBox的bug
    element-ui上传文件带token
    MySQL启动出现The server quit without updating PID file错误解决办法
    重装应用商店
    vscode在vue-cli中按照ESlint自动格式化代码
  • 原文地址:https://www.cnblogs.com/13224ACMer/p/4728953.html
Copyright © 2011-2022 走看看