11078 - Open Credit System

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 anycourse they like, but there is a problem. Some of the students are more seniorthan other students. The professor of such a course has found quite a number ofsuch students who came from senior classes (as if they came to attend the prerequisite course after passing an advanced course). But he wants to do justiceto the new students. So, he is going to take a placement test (basically an IQtest) to assess the level of difference among the students. He wants to knowthe maximum amount of score that a senior student gets more than any juniorstudent. 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. Helpthe professor to figure out a solution.

Input
Input consists of a number of test cases T (less than 20). Each casestarts with an integer n which is the number of students in thecourse. This value can be as large as 100,000 and as low as 2. Next n linescontain n integers where the i'th integer is thescore of the i'th student. All these integers have absolutevalues less than 150000. If i < j, then i'th studentis senior to the j'th student.

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

SampleInput                             Outputfor Sample Input

3

2

100

20

4

4

3

2

1

4

1

2

3

4

 

80
3
-1

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。。。

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）；

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

 

 

#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;}