hdu1003 Max Sum （求连续子区间最大和）

Max Sum

                                                                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 169481    Accepted Submission(s): 39557

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

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   解析：求连续子区间最大和。

               用f【i】表示以第 i 个数字结尾的连续子区间的最大值，则 f【i】=max{f【i-1】+a【i】，a【i】}，然后分情况处理一下起点、终点，并更新当前最优值即可。

代码：

#include<cstdio>
#define inf 1000000000
#define maxn 1000
using namespace std;

int ans,ans_l,ans_r;

void work()
{
  int i,j,k,n,x,sum,l,r;
  scanf("%d",&n);
  ans=-inf,sum=0,l=1,r=0;
  for(i=1;i<=n;i++)
    {
      scanf("%d",&x);
      if(sum<0)sum=x,l=r=i;
      else sum+=x,r=i;
      
    if(sum>ans)ans=sum,ans_l=l,ans_r=r;  
    }
  printf("%d %d %d\n",ans,ans_l,ans_r);  
}

int main()
{
  freopen("1.in","r",stdin);
  int t,i;
  while(scanf("%d",&t)!=EOF)
    for(i=1;i<=t;i++)
      {
        printf("Case %d:\n",i);
        work();
        if(i<t)printf("\n");
      }
  return 0;
}