HDU 1003 MAX SUM java 实现

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 141614 Accepted Submission(s): 32953

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
/* * 典型动态规划的思想 动态规划的经典题   之前我一直不是很理解动态规划    * 我个人做题的时候 遇到动态规划的题型 直接用递归   * 算法分析：求最大字段和，d[i]表示在已i结尾（包含i）在a[1...i]上的最大和，d[i]=(d[i-1]+a[i]>)?d[i-1]+a[i]:a[i]; * max={d[i],n>=i>=1} *  * */import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner in =new Scanner(System.in);final int N=100010; //注意定义多一些  开始我定义少了 提交三次都是WAint[] a =new int[N];int[] d=new int[N];int ee=0;    int test=in.nextInt();    for(int j=0;j<test;j++){    int x=in.nextInt();    for(int y=1;y<=x;y++){    a[y]=in.nextInt();    }    d[1]=a[1];    for(int m=2;m<=x;m++)// 动态规划的核心     {    if(d[m-1]<0)    d[m]=a[m];    else    d[m]=d[m-1]+a[m];    }    int max=d[1],e=1;    for(int a1=2;a1<x+1;a1++)//找出最右边的点    {        if(max<d[a1])    {    max=d[a1];    e=a1;    }    }    int t=0,f=e;    for(int i=e;i>0;i--)//找出最左边的点    {    t=t+a[i];    if(t==max)    f=i;    }    System.out.println("Case "+(j+1)+":");    System.out.println(max+" "+f+" "+e);    ee++;    if(ee!=test)    System.out.println();    }}}

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