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;
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 }