Open Credit System（UVA11078）

11078 - Open Credit System Time limit: 3.000 seconds

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

80 3 -1

[b]题意：给一个长度为n的整数序列a0,a1,a2,,,,,,an-1,找出两个整数ai和aj（i<j）使得ai-aj最大。。。[/b]

[b]思路：如果直接二重循环是不可取的，因为时间复杂度O（n^2）在n=100000下会超时，所以我们可以选择小于j的最大ai，每次记录ans，最后的ans就是结果，优化后可以是、使时间和空间复杂度都变成O（n）；[/b]

[b]ps：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2019[/b]

#include<cstdio>
#include<algorithm>
using namespace std;
int num[100005];
int maxnum,ans;

int main()
{
int T,N,i,j;
scanf("%d",&T);
while(T--)
{
scanf("%d",&N);
scanf("%d%d",&num[0],&num[1]);
maxnum=num[0]>num[1]?num[0]:num[1];
ans=num[0]-num[1];
for(i=2;i<N;i++)
{
scanf("%d",&num[i]);
ans=max(ans,maxnum-num[i]);
maxnum=max(maxnum,num[i]);
}
printf("%d\n",ans);
}
return 0;
}