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A - Class Statistics
Time Limit:3000MS
Memory Limit:0KB
64bit IO Format:%lld & %lluSubmit Status Practice UVALive 5682
Description
///最開始 沒有讀懂 Largest gap坑也~~~ 是拍好循序後的 連續最大差 #include <iostream> #include <string.h> #include <stdio.h> #include <algorithm> using namespace std; int main() { int t; int num; int n;int a[150]; scanf("%d",&t); for(num=1;num<=t;num++) { scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d",&a[i]); sort(a,a+n); int sum=0; for(int i=0;i<n-1;i++) { if(sum<(a[i+1]-a[i])) sum=(a[i+1]-a[i]); } printf("Class %d\n",num); printf("Max %d, Min %d, Largest gap %d\n",a[n-1],a[0],sum); } return 0; }
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 ( 1K100) 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 ( 2N50) 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
25 30 25 76 23 786 25 50 70 99 70 90
Sample Output
Class 1Max 78, Min 23, Largest gap 46Class 2Max 99, Min 25, Largest gap 25