HDU–1003 Max Sum（最大连续子序列和）

Problem DescriptionGiven 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. InputThe 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). OutputFor 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 Input25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5 Sample OutputCase 1: 14 1 4 Case 2: 7 1 6 AuthorIgnatius.L RecommendWe have carefully selected several similar problems for you:  1176 1069 2084 1058 1203 代码：

 #include<cstdio>
 int a[+];
 int main()
 {
     int t,n,k=;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         for(int i=;i<=n;i++)
             scanf("%d",&a[i]);
         int max=-,st,endd,sum=,st1=; //注意整数的范围
         for(int i=;i<=n;i++)
         {
             sum+=a[i];
             if(sum>max)
             {
                 max=sum;
                 st=st1;
                 endd=i;
             }
             if(sum<)
             {
                 sum=;
                 st1=i+;    //st1是临时的开始点，如果后面的sum<0，这个开始点也就不会计入
             }
         }
         printf("Case %d:\n%d %d %d\n",k++,max,st,endd);
         if(t!=)
         {
             printf("\n");
         }     }
     return ;
 }