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.
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).
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.
和最大的最长子序列：
#include<stdio.h>
int main ()
{
int T, i, n, x, a, b, c, sum, M, k = 0, flag = 0;
scanf("%d", &T);
while (T--)
{
k++;
scanf("%d %d", &n, &x);
sum = M = x;
a = b = c = 1;
for (i = 2; i <= n; i++)
{
scanf("%d", &x);
if (sum + x < x)
{
sum = x;
a = i;
} //若sum+x比x小，则将sum置为x，暂将此时的位置记录下来
else
sum += x; //否则直接加上
if (sum > M)
{
M = sum;
b = a;
c = i;
} //当找到更大子序列的和时更改最大值和对应位置
}
if (flag == 1)
printf("\n");
printf("Case %d:\n", k);
printf("%d %d %d\n", M, b, c);
flag = 1;
}
return 0;
}
