hdu1003 Max Sum--DP

原题链接：​​http://acm.hdu.edu.cn/showproblem.php?pid=1003​​

一：原题内容

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

二：分析理解

一个简单动态规划。。。。

三：AC代码

#include<iostream>
#include<string.h>
#include<algorithm>

using namespace std;

int a[100010];

int main()
{
  int T;
  scanf("%d", &T);

  for (int i = 1; i <= T; i++)
  {
    scanf("%d", &a[0]);

    for (int j = 1; j <= a[0]; j++)
      scanf("%d", &a[j]);

    int left = 1;
    int right = 1;
    int cnt = a[1];
    int flag = 1;
    int max = a[1];

    for (int k = 2; k <= a[0]; k++)
    {
      if (cnt < 0)
      {
        flag = k;
        cnt = a[k];
      }
      else
        cnt += a[k];

      if (cnt > max)
      {
        left = flag;
        right = k;
        max = cnt;
      }
    }

    printf("Case %d:\n%d %d %d\n", i, max, left, right);
    if (i != T)
      printf("\n");
  }

  return 0;
}