# HDOJ 1003 Max Sum

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

```import java.util.Scanner;

public class Main {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
int time = 1;
while(t-->0){
int m = sc.nextInt();
int p1,p2=1;
int n = sc.nextInt();
int now=n;
int max=n;
int x=1;
p1=1;
for(int i=1;i<m;i++){
n=sc.nextInt();
if(n>n+now){
x=i+1;
now=n;
}else{
now = n+now;
}
if(now>max){
max=now;
p1=x;
p2=i+1;
}

}

System.out.println("Case "+time+":");
time++;
System.out.println(max+" "+p1+" "+p2);
if(t!=0){
System.out.println();
}

}

}

}

```
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