题目描述:
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 321851 Accepted Submission(s): 76533
Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
Sample Input
2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5
Sample Output
Case 1: 14 1 4 Case 2: 7 1 6
Java 代码实现，其中有个小坑，就是dp数组要开的大些，不然一直WA，不知道错误在哪里
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
int[] array = new int[100010];
int[] dp = new int[100010];
int t = cin.nextInt();
for (int count = 0; count < t; count++) {
int n = cin.nextInt();
for(int i = 0;i<n;i++){
array[i] = cin.nextInt();
}
dp[0] = array[0];
for(int i =1;i<n;i++){
dp[i] = Math.max(dp[i-1]+array[i], array[i]);
}
int max = dp[0];
int endIndex = 0;
for(int i = 1;i<n;i++){
if(dp[i]>max){
max = dp[i];
endIndex = i;
}
}
int temp =0,l = endIndex;
for(int i = endIndex;i>=0;i--){
temp+=array[i];
if(temp==max){
l = i;
}
}
if(count!=0){
System.out.println();
}
System.out.println("Case "+(count+1)+":");
System.out.println(max+" "+(l+1)+" "+(endIndex+1));
}
}
}