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'thstudent.
All these integers have absolute values less than 150000. If i < j, then i'thstudent 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 |
这道题，就是给出一系列的有序的数，比如a[0] a[1],a[2],a[3].....a[i]....a[n]. 要你求出a[i]-a[j]的最大值（注意i<j ）
而且数据规模也是十万加的.....，可想而知，朴树的算法是会超时..
采用动态规划的思想...
先逐渐求出区间的最大值，注意每一次扩展时，都进行一次比较...
代码如下：
1 #include<cstdio> 2 #include<cstring> 3 const int maxn=100000; 4 int aa[maxn]; 5 inline int max(int a,int b){ 6 return a>b?a:b; 7 } 8 9 int main(){ 10 int n,m; 11 //freopen("test.in","r",stdin); 12 scanf("%d",&n); 13 while(n--){ 14 scanf("%d",&m); 15 for(int i=0;i<m;i++) 16 scanf("%d",aa+i); 17 int cnt=aa[0]-aa[1]; 18 int maxc=aa[0]; 19 for(int i=1;i<m;i++){ 20 cnt=max(cnt,maxc-aa[i]); 21 maxc=max(maxc,aa[i]); 22 } 23 printf("%d ",cnt); 24 } 25 return 0; 26 }
进一步优化之后的
代码：
1 #include<cstdio> 2 #include<cstring> 3 const int maxn=100000; 4 inline int max(int a,int b){ 5 return a>b?a:b; 6 } 7 int main(){ 8 int n,m; 9 int b,c; 10 int ans; 11 freopen("test.in","r",stdin); 12 scanf("%d",&n); 13 while(n--){ 14 scanf("%d",&m); 15 scanf("%d",&b); 16 ans =-200000; 17 for(int i=1;i<m;i++){ 18 scanf("%d",&c); 19 ans=max(ans,b-c); 20 b=max(b,c); 21 } 22 printf("%d ",ans); 23 } 24 return 0; 25 }