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）

 1 #include<stdio.h>
 2 
 3 struct Subsq
 4 { int sum; int l; int r; };
 5 
 6 struct Subsq max_sum(int a[], int left, int right)    // 在区间[left，right)寻找最大连续和 
 7 {
 8     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 
 9     int mid, i;
10     b.l=left; b.r=right;
11     if(b.r-b.l==1) b.sum=a[b.l];    // 若只有一个元素，则返回它 
12     else
13     {
14         mid=b.l+(b.r-b.l)/2;
15         leftsq=max_sum(a,b.l,mid); rightsq=max_sum(a,mid,b.r);
16         int sum1=a[mid-1], sum2=a[mid], ls=0, rs=0, l=mid-1, r=mid+1;    // 起点在中间，分别向左、右推进 
17         for(i=mid-1;i>=b.l;i--)
18         {
19             ls+=a[i];
20             if(ls>=sum1)
21             { sum1=ls; l=i; }
22         }
23         for(i=mid;i<b.r;i++)
24         {
25             rs+=a[i];
26             if(rs>sum2)
27             { sum2=rs; r=i+1; }
28         }
29         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 
30         if(b.sum<=leftsq.sum)    // If there are more than one result, output the first one.
31         { b.sum=leftsq.sum; b.l=leftsq.l; b.r=leftsq.r; }
32         if(b.sum<rightsq.sum)
33         { b.sum=rightsq.sum; b.l=rightsq.l; b.r=rightsq.r; }
34     }
35     return b;
36 }
37 
38 int main()
39 {
40     struct Subsq b;
41     int t, n, a[100000], i, j;
42     scanf("%d", &t);
43     for(j=1;j<=t;j++)
44     {
45         scanf("%d", &n);
46         for(i=0;i<n;i++)
47             scanf("%d", &a[i]);
48         b=max_sum(a,0,n);
49         printf("Case %d:\n%d %d %d\n", j, b.sum, b.l+1, b.r);
50         if(j<t) printf("\n");
51     }
52     return 0;
53 }

AC

// 补充：最大连续和问题（详见紫书8.1）