Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 128098 Accepted Submission(s): 29672
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
Author
Ignatius.L
Recommend
又来水题了
2bb的错误，想不出的状态方程。。。
import java.util.Scanner;
public class Main {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
for(int i=0; i<T; i++){
int n = sc.nextInt();
int[] dp = new int[n+1], a = new int[n+1];
for(int j=0; j<n; j++) a[j]=sc.nextInt();
int maxSum=a[0], right=0, left=0;
for(int j=0; j<n; j++){
dp[j]=Math.max((j==0?0:dp[j-1]), 0)+a[j];
if(dp[j]>maxSum){
maxSum=dp[j]; right=j;
}
}
for(int j=right; j>-1; j--){
dp[j]=Math.max((j==right?0:dp[j+1]), 0)+a[j];
if(maxSum==dp[j])
left=j;
}
if(i>0) System.out.println();
System.out.println("Case "+(i+1)+":");
System.out.println(maxSum+" "+(++left)+" "+(++right));
}
}
}