HDOJ 1003

HDOJ 1003

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.

code

import java.math.BigInteger;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int T = in.nextInt();
        for(int i1 = 0; i1 < T; i1++) {
            int N = in.nextInt();
            int a[] = new int[N];
            int sum = 0;
            int max = -1001;
            int start = 0;
            int end = 0;
            int temp = 0;
            for(int i2 = 0; i2 < N; i2++) {
                a[i2] = in.nextInt();
                sum += a[i2];
                if(sum > max) {
                    max = sum;
                    start = temp;
                    end = i2;
                }
                if(sum < 0) {
                    sum = 0;
                    temp = i2 + 1;
                }
            }
            System.out.println("Case "+(i1+1)+":");
            System.out.println(max+" "+(start+1)+" "+(end+1));
            if(i1 != T-1) {
                System.out.println();
            }
        }
    }
    
}

tips

对于一个序列，求最大字序列，初始子序列首尾都为0，每相加下一个数，若和sum大于max，则更新最大子序列的首尾，尾为当前数字的序号。若sum小于0，说明这段子序列的和为负，需要重新开始计算另一端子序列，所以sum重置为0，用temp+1标记为另一段子序列的首。