7A - Max Sum

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

// WA*9，Time Limit Exceeded*2
代码省略
// 当需要函数返回多个值时，可使用结构类型
// 分治法（详见紫书8.1.3）（注意点见代码）

 #include<stdio.h>

 struct Subsq
 { int sum; int l; int r; };

 struct Subsq max_sum(int a[], int left, int right)    // 在区间[left，right)寻找最大连续和
 {
     struct Subsq b, leftsq, rightsq;    // 最优解要么全在左半边，要么全在右半边，要么起点在左半边、终点在右半边
     int mid, i;
     b.l=left; b.r=right;
     if(b.r-b.l==) b.sum=a[b.l];    // 若只有一个元素，则返回它
     else
     {
         mid=b.l+(b.r-b.l)/;
         leftsq=max_sum(a,b.l,mid); rightsq=max_sum(a,mid,b.r);
         int sum1=a[mid-], sum2=a[mid], ls=, rs=, l=mid-, r=mid+;    // 起点在中间，分别向左、右推进
         for(i=mid-;i>=b.l;i--)
         {
             ls+=a[i];
             if(ls>=sum1)
             { sum1=ls; l=i; }
         }
         for(i=mid;i<b.r;i++)
         {
             rs+=a[i];
             if(rs>sum2)
             { sum2=rs; r=i+; }
         }
         b.sum=sum1+sum2; b.l=l; b.r=r;    // 记录起点在左半边、终点在右半边情况下的最大连续和
         if(b.sum<=leftsq.sum)    // If there are more than one result, output the first one.
         { b.sum=leftsq.sum; b.l=leftsq.l; b.r=leftsq.r; }
         if(b.sum<rightsq.sum)
         { b.sum=rightsq.sum; b.l=rightsq.l; b.r=rightsq.r; }
     }
     return b;
 }

 int main()
 {
     struct Subsq b;
     int t, n, a[], i, j;
     scanf("%d", &t);
     for(j=;j<=t;j++)
     {
         scanf("%d", &n);
         for(i=;i<n;i++)
             scanf("%d", &a[i]);
         b=max_sum(a,,n);
         printf("Case %d:\n%d %d %d\n", j, b.sum, b.l+, b.r);
         if(j<t) printf("\n");
     }
     return ;
 }

AC

// 补充：最大连续和问题（详见紫书8.1）