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
我的代码
#include
int main()
{
int i,t,n,a[100000],b[3],k=1,f,s,min,sum;
scanf("%d",&t);
if(t>=1&&t<=20)
while(t--)
{
scanf("%d",&n);min=0;sum=0;
if(1<=n&&n
{
for(i=0;i
{
while(1)
{
scanf("%d",&a[i]);
if(a[i]>=-1000&&a[i]
}
if(min>a[i])min=a[i];
}
for(i=0;i
if(a[i]!=min)
sum+=a[i];
for(i=0;i
if(a[i]>=0)
{f=i+1;break;}
for(i=n-1;i>=0;i--)
if(a[i]>=0)
{s=i+1;break;}
printf("Case %d:\n",k++);
printf("%d %d %d\n",sum,f,s);
if(t>=1)printf("\n");
}
}
return 0;
}
