J - Max Sum

J - Max Sum

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

//第一行代表有 t 组测试案例，问最大连续的和为多少，并输出第几位数到第几位数
//虽然，这是一道dp水题，但我从没做过，自己写出来感觉不错，62ms 比别人的慢了一倍，应该是我用了两次一重循环吧。

 #include <stdio.h>

 #define MAXN 100005

 int num[MAXN];
 int dp[MAXN];

 int cmp(int a,int b)
 {
     return a>b?a:b;
 }

 int main()
 {
     int t;
     int n,i,Case=;
     scanf("%d",&t);
     while (t--)
     {
         scanf("%d",&n);
         int max;
         for (i=;i<=n;i++)
         {
             if (i==)
             {
                 scanf("%d",&num[i]);
                 dp[i]=num[i];
                 max=i;
                 continue;
             }
             scanf("%d",&num[i]);
             dp[i]=cmp(num[i],dp[i-]+num[i]);
             if (dp[i]>dp[max]) max=i;
         }
         int s=max;
         for (i=max;i>=;i--)
         {
             if (dp[i]>=)
                 s=i;
             if (dp[i]<)
                 break;
         }
         printf("Case %d:\n",++Case);
         printf("%d %d %d\n",dp[max],s,max);
         if (t!=) printf("\n");
     }
     return ;
 }