（HDOJ 1003）Max Sum

Max Sum[align=left]Problem Description[/align]Given a sequence a[1],a[2],a[3]......a
, 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.
[align=left]Input[/align]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).
[align=left]Output[/align]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.
[align=left]Sample Input[/align]2 5 6 -1 5 4 -7

7 0 6 -1 1 -6 7 -5

[align=left]Sample Output[/align]Case 1: 14 1 4 Case 2: 7 1 6

[align=left]Author[/align]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;
scanf("%d",&n);
while(n--)
{
scanf("%d",&num);
a = (int *)calloc(num,sizeof(int));
for(i = 0; i < num ;i ++)
scanf("%d",&a[i]);
for( l = 0,r = 0,sum = 0,max = a[0],i = 0;i <num ;i ++)
{
for(sum = 0,j = i ;j <num ;j ++)
{
sum += a[j];
if(sum > max)
{
max = sum ;
l = i;
r = j;
}
if(sum < 0)
{
i = j;
sum =0;
break;
}
}
}
printf("Case %d:\n%d %d %d\n",t++,max ,l+1 ,r+1);
if( n)
printf("\n");
//getchar();
}
return 0; }