Open Credit System - UVa 11078 水题

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本博客介绍了一个开放学分系统中教授如何通过测试评估高级与初级学生之间的评分差距，并提供了一段AC代码实现该功能。输入包括多个测试案例，每个案例由一系列学生的分数组成，高级学生在前。

Open Credit System
Input: Standard Input

Output: Standard Output

In an open credit system, the students can choose any course they like, but there is a problem. Some of the students are more senior than other students. The professor of such a course has found quite a number of such students who came from senior classes (as if they came to attend the pre requisite course after passing an advanced course). But he wants to do justice to the new students. So, he is going to take a placement test (basically an IQ test) to assess the level of difference among the students. He wants to know the maximum amount of score that a senior student gets more than any junior student. For example, if a senior student gets 80 and a junior student gets 70, then this amount is 10. Be careful that we don't want the absolute value. Help the professor to figure out a solution.

Input
Input consists of a number of test cases T (less than 20). Each case starts with an integer n which is the number of students in the course. This value can be as large as 100,000 and as low as 2. Next n lines contain n integers where the i'th integer is the score of the i'th student. All these integers have absolute values less than 150000. If i < j, then i'th student is senior to the j'th student.

Output
For each test case, output the desired number in a new line. Follow the format shown in sample input-output section.

Sample Input Output for Sample Input

80
3
-1

Problemsetter: Mohammad Sajjad Hossain

Special Thanks: Shahriar Manzoor

题意：前面的数是高年级的，找到高年级得分比低年级的差的最大值。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
typedef long long ll;
using namespace std;
int num[100000010],num2[100000010];
int main()
{ int t,n,i,j,k,ans;
  scanf("%d",&t);
  while(t--)
  { ans=-1000000000;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    { scanf("%d",&num[i]);
      num2[i]=-1;
    }
    num2[n]=num[n];
    for(i=n-1;i>=2;i--)
     num2[i]=min(num2[i+1],num[i]);
    for(i=1;i<=n-1;i++)
        ans=max(ans,num[i]-num2[i+1]);
    printf("%d\n",ans);
  }
}