（HDOJ 1003）Max Sum

Max Sum

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

 

AC code:

 1 #include<stdio.h> 
 2 #include<string.h> 
 3 #include<stdlib.h>   
 4 int main() 
 5 { 
 6   
 7     int r = 0,l = 0,i= 0 ,j = 0,num = 0,n;// l用来记录最大左范围r 右 
 8     int *a;//,[6]={5,6,5,-4,-7,3}; 
 9     int sum = 0,max = 0,t= 1; 
10     scanf("%d",&n); 
11     while(n--) 
12     { 
13         scanf("%d",&num); 
14         a = (int *)calloc(num,sizeof(int)); 
15         for(i = 0; i < num ;i ++) 
16             scanf("%d",&a[i]); 
17         for( l = 0,r = 0,sum = 0,max = a[0],i = 0;i <num ;i ++) 
18         { 
19             for(sum = 0,j = i ;j <num ;j ++) 
20             { 
21                 sum += a[j]; 
22                 if(sum > max) 
23                 { 
24                     max = sum ; 
25                     l = i; 
26                     r = j; 
27                 } 
28                 if(sum < 0) 
29                 { 
30                     i = j; 
31                     sum =0; 
32                     break; 
33                 } 
34             } 
35         } 
36         printf("Case %d:\n%d %d %d\n",t++,max ,l+1 ,r+1); 
37         if( n) 
38             printf("\n"); 
39         //getchar(); 
40     } 
41     return 0; 

42 }