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
题意描述：
输入一个T表示T组样例，然后输入n表示有n个数，求连续的最大数的和，可以先定义一个sum，表示连续数字的和，如果sum小于0了，则把sum初始化为0，然后再定义一个max，存储最大sum，也就是最大连续数字的和。
程序代码：
#include<stdio.h>
int a[100010];
int main()
{
int T,i,k,n,sum,max,p,q,t;
scanf("%d",&T);
for(k=1;k<=T;k++)
{
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
sum=0;
max=-99999999;
t=p=q=1;
for(i=1;i<=n;i++)
{
sum+=a[i];
if(sum>max)
{
max=sum;
p=t;
q=i;
}
if(sum<0)
{
sum=0;
t=i+1;
}
}
printf("Case %d:\n%d %d %d\n",k,max,p,q);
if(k!=T)
printf("\n");
}
return 0;
}