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
#include
#include
using namespace std;
const int maxn=100000+10;
int main()
{
int n,T,a[maxn],max,begin,end,k,i,t,j;
scanf("%d",&n);
for (j=1;j<=n;j++)
{
t=0;max=-9999;k=1;//max的初始值很重要，因为子串中可能有负数
scanf("%d",&T);
for (i=1;i<=T;i++)
{
scanf("%d",a+i);
t+=a[i];
if (t>max)//找到更大的一个子串，更新最大值和起点，终点
{
max=t;
begin=k;
end=i;
}
if (t<0)//可能还有比之前大的子串
{
t=0;
k=i+1;//不能把i+1赋值给begin，因为不一定下个子串和最大
}
}
printf("Case %d:\n",j);
if (j!=n)
{
printf("%d %d %d\n\n",max,begin,end);
}
else
printf("%d %d %d\n",max,begin,end);
}
return 0;
}
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