[动态规划]hdu 1003 Max Sum 最大子列

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 242466 Accepted Submission(s): 57246

Problem Description

Given a sequence a[1],a[2],a[3]……a
, 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

package hdu;

import java.util.Scanner;

public class MaxSum {

public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t = sc.nextInt();
int n,i,j,u,max;
int[] sum;
int[] a;
int[] b;
for(u=0 ;u<t ;u++){
n = sc.nextInt();
a=new int
;
for(i=0;i<n;i++)
a[i]=sc.nextInt();
max=0;j=0;
sum=new int
;
b=new int
;
sum[0]=a[0];
max=sum[0];
b[0]=0;
for(i=1;i<n;i++){
if(a[i]>a[i]+sum[i-1]){
sum[i]=a[i];
b[i]=i;
}
else{
sum[i]=sum[i-1]+a[i];
b[i]=b[i-1];
}
if(sum[i]>max){
max=sum[i];j=i;
}
}
System.out.println("Case "+(u+1)+":");
System.out.println(max+" "+(b[j]+1)+" "+(j+1));
if(u!=t-1)
System.out.println();
}
}
}