UVA - 11078 Open Credit System

Open Credit System

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

Problem E
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 nintegers 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

3 2 100 20 4 4 3 2 1 4 1 2 3 4

80 3 -1

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Problemsetter: Mohammad SajjadHossain

Special Thanks: Shahriar Manzoor

 

分析：

刘汝佳大白书里的题，简单题。唯一需要注意的地方是：如何把两重循环降为一重循环。

ac代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=100000;
int a[maxn],n;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
        int ans=a[0]-a[1];
        int maxai=a[0];
        for(int j=1;j<n;j++)//j从1开始枚举，因为j=0时不存在
        {
            ans=max(ans,maxai-a[j]);
            maxai=max(a[j],maxai);
        }
        printf("%d\n",ans);
    }
    return 0;
}