【题目】

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 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

3
2
100
20
4
4
3
2
1

4

1

2

3

4

Problemsetter: Mohammad Sajjad Hossain

Special Thanks: Shahriar Manzoor

【分析】

【思路一】

最简单的一种就是二重循环。但是Time limit exceeded。         这种算法的时间复杂度为O（n^2）。在n 很大规模的时候就无能为力了。

【思路二】

对于每个固定的j，我们应该选择的是小于j且Ai最大的i，而和Aj的具体数值无关。这样，我们从小到大枚举j,顺便维护Ai的最大值即可。时间复杂度降为O（n）

【代码】

【代码1】

/*********************************
*   日期：2014-5-1
*   作者：SJF0115
*   题号: 11078 - Open Credit System
*   地址：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=22&page=show_problem&problem=2019
*   来源：UVA
*   结果：Time limit exceeded。
*   总结：
**********************************/
#include <iostream>
#include <stdio.h>
using namespace std;

int A[100001];

int main(){
    int T,i,j,n;
    scanf("%d",&T);
    //T组测试数据
    while(T--){
        scanf("%d",&n);
        //输入数据
        for(i = 0;i < n;i++){
            scanf("%d",&A[i]);
        }
        //初始化 不要初始化为0因为最终答案可能小于0
        int max = A[0] - A[1];
        //计算最大的A[i]-A[j]
        for(i = 0;i < n;i++){
            for(j = i+1;j < n;j++){
                if(max < A[i] - A[j]){
                    max = A[i] - A[j];
                }//if
            }//for
        }//for
        printf("%d\n",max);
    }
    return 0;
}

/*********************************
*   日期：2014-5-1
*   作者：SJF0115
*   题号: 11078 - Open Credit System
*   地址：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=22&page=show_problem&problem=2019
*   来源：UVA
*   结果：Accepted
*   总结：
**********************************/
#include <iostream>
#include <stdio.h>
using namespace std;

int A[100001];

int main(){
    int T,i,j,n;
    scanf("%d",&T);
    //T组测试数据
    while(T--){
        scanf("%d",&n);
        //输入数据
        for(i = 0;i < n;i++){
            scanf("%d",&A[i]);
        }
        //初始化 不要初始化为0因为最终答案可能小于0
        int max = A[0] - A[1];
        //maxAi动态维护A[0],A[1]....A[j-1]的最大值
        int maxAi = A[0];
        //计算最大的A[i]-A[j]
        for(i = 1;i < n;i++){
            //更新最大A[i]-A[j]
            if(max < maxAi - A[i]){
                max = maxAi - A[i];
            }
            //更新最大元素
            if(maxAi < A[i]){
                maxAi = A[i];
            }
        }//for
        printf("%d\n",max);
    }
    return 0;
}