杭电 1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 172675    Accepted Submission(s): 40223

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
  
  

 

分析：

 

这是一道求最大子段和的问题.

定义一个初始值为-1001的max（测试数据最小为-1000）从第一个数开始累加sum，

每次累加之后与max打擂，若大于则更新子段的终止位置。

若sum<0则丢弃之前的子段，从终止位置的下一位置重新开始；

直到再找到一个更大的子段，更新起始位置；

 

第一次做的时候在sum<0时直接就更新了起始位置到当前位置的下一位置，导致WA。

（应该确保之后找到了更大的子段和时才更新起始位置。）因此纠结了好半天；

 

 

代码：

 

#include<iostream>

using namespace std;

int main()

{

int k;

cin >> k;

for (int j = 1;j <= k;j++)

{

int begin = 1, end, begin_t=1;

int sum = 0, max = -1001,shuru;

int n;

int i;

cin >> n;

for (i = 1;i <= n;i++)

{

scanf_s("%d", &shuru);

sum += shuru;

if (sum > max)

{

max = sum;

end = i;

begin = begin_t;

}

if (sum < 0)

{

sum = 0;

begin_t = i + 1;

}

}

printf("Case %d:\n", j);

printf(j==k ? "%d %d %d\n" : "%d %d %d\n\n", max, begin, end);

}

return 0;

}