218
UVA之11078 - Open Credit System
【題目】
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 |
80 |
Problemsetter: Mohammad Sajjad Hossain
Special Thanks: Shahriar Manzoor
【分析】
【思路一】
最簡單的一種就是二重循環。但是Time limit exceeded。 這種算法的時間複雜度為O（n^2）。在n 很大規模的時候就無能為力了。
【思路二】
對於每個固定的j，我們應該選擇的是小於j且Ai最大的i，而和Aj的具體數值無關。這樣，我們從小到大枚舉j,順便維護Ai的最大值即可。時間複雜度降為O（n）
【代碼】
【代碼1】
/*********************************
* 日期：2014-5-1
* 作者：SJF0115
* 題號: 11078 - Open Credit System
* 地址：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=22&page=show_problem&problem=2019
* 來源：UVA
* 結果：Time limit exceeded。
* 總結：
**********************************/
#include <iostream>
#include <stdio.h>
using namespace std;
int A[100001];
int main(){
int T,i,j,n;
scanf("%d",&T);
//T組測試數據
while(T--){
scanf("%d",&n);
//輸入數據
for(i = 0;i < n;i++){
scanf("%d",&A[i]);
}
//初始化 不要初始化為0因為最終答案可能小於0
int max = A[0] - A[1];
//計算最大的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;
}
【代碼2】
/*********************************
* 日期：2014-5-1
* 作者：SJF0115
* 題號: 11078 - Open Credit System
* 地址：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=22&page=show_problem&problem=2019
* 來源：UVA
* 結果：Accepted
* 總結：
**********************************/
#include <iostream>
#include <stdio.h>
using namespace std;
int A[100001];
int main(){
int T,i,j,n;
scanf("%d",&T);
//T組測試數據
while(T--){
scanf("%d",&n);
//輸入數據
for(i = 0;i < n;i++){
scanf("%d",&A[i]);
}
//初始化 不要初始化為0因為最終答案可能小於0
int max = A[0] - A[1];
//maxAi動態維護A[0],A[1]....A[j-1]的最大值
int maxAi = A[0];
//計算最大的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;
}
最後更新：2017-04-03 12:56:27