杭电ACM OJ 1003 Max Sum 一点点的动态规划思想 入门级

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 260585    Accepted Submission(s): 61936

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6
翻译：就是几个int数，按序拜访 如6 -1 5 4 -7 找到其中连续的几个数 使得和为最大（不一定要从第一个数头开始）
第一个数是代表这个数组有几个 后面是具体的数
public class MaxSum1003 {    private static List<Integer> list;    private static int begin = 0;    private static int end = 0;    private static int max = 0;    public static void main(final String[] args) throws Exception {        initData();        calculate(list);        System.out.println("" + max + (begin+1) + (end+1));    }    private static void initData() {        list = new ArrayList<>();        list.add(6);        list.add(-1);        list.add(5);        list.add(4);        list.add(-7);    }    private static void calculate(List<Integer> list) {        int size = list.size();        int sum = 0;        //如果固定从头开始，直接开始累加，不断和最大值比较//        for (int i = 0; i < size; i++) {//            sum += list.get(i);//            max = Math.max(sum, max);//        }        //如果不一定要从头开始，就是本题情况，从一串数中，截取一段，需要是和最大        //这里和上面唯一的区别是需要加个if来判断，是否舍弃前面的东西        for (int i = 0; i < size; i++) {            sum += list.get(i);            if (sum > max) {                max = sum;                end = i;            }            //如果前面的sum和小于0了，还要他干什么，所以舍弃            if (sum < 0) {                sum = 0;                begin = i;            }        }    }}