Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 81597 Accepted Submission(s): 18767

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

问题描述：取一个数组中连续的m个数，使其总和最大

算法分析：DP思路，此题不仅要求输出最大的和而且要求输出他的起始位置和结束位置。我的思路是定义状态temp[]数组，temp[i]表示以i作为尾巴的总和最大的连续数的和，这里的temp[i]可能是i一个数字，也可能是i加上i前面的连续几个数字。

实现：首先temp[1] = input[1] ，px[1] = 1 py[1] = 1 对于temp[i](i >= 2) ，如果temp[i-1]>=0 在i-1的基础上加上i必然是temp[i]的最大值，则temp[i] = temp[i-1] + input[i] ，px[i] = px[i-1] py[i] = i ;如果temp[i-1] < 0 则temp[i] 只能等于input[i]否则会更小。

所以temp[i] = input[i] + temp[i-1]>=0?temp[i-1]:0 ；即是新的状态方程

代码如下：

/*Memory: 1768 KB   Time: 31 MS  
Language: GCC   Result: Accepted  
 
This source is shared by hust_lcl
*/
#include <stdio.h>
#include <stdlib.h>
int input[100010];
int temp[100010];
int px[100010];
int py[100010];
int main()
{
    int n , m , i , flag , k = 1 , j;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&m);
        for(i = 1 ; i <= m ; i ++)
          scanf("%d",&input[i]);
        printf("Case %d:\n",k++);
        temp[1] = input[1];
        px[1] = 1;
        py[1] = 1;
        for(i = 2 ; i <= m ; i ++ )
        {
            if(temp[i-1]>=0)
              {
                  temp[i] = input[i] + temp[i-1];
                  px[i] = px[i-1];
                  py[i] = i;
              }
            else
              {
                  temp[i] = input[i];
                  px[i] = i ;
                  py[i] = i;
              }
        }
        flag = temp[1];
        j = 1;
        for(i = 1 ; i <= m ; i++)
          if(temp[i] > flag)
          {
              flag = temp[i];
              j = i;
          }
        printf("%d %d %d\n",flag, px[j],py[j]);
        if(n>0) printf("\n");
    }
    return 0;
}

DP课题2 HDOJ 1003 Max Sum

Max Sum

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