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<stdio.h>
int main()
{
int end,begin,temp,max;
int e,f,k=0,flag=1;
int a[100005],n,m,i; //数组接收数据,次数,个数,下标变量
//初始化
scanf("%d",&n); //输入次数
while(n--)
{
begin=end=0;
k++; //case次数
scanf("%d",&m); //输入数据
for(i=1;i<=m;i++)
scanf("%d",&a[i]);
temp=0; // 和初始化
max=-999999; //存放最小值
begin=e=1; //标记输入数据后,begin,e变化
f=1; //初始化
for(i=1;i<=m;i++)
{
temp +=a[i];
if(temp>max) //大于最小值,把变量赋值给全新的变量,最后输出时不会出问题
{
e=begin;
f=i;
max=temp;
}
if(temp<0) //本人有考虑到
{
begin = i+1; //当和小于0时,跳过,归零重新计算
temp = 0;
}
}
printf("Case %d:\n",k);
printf("%d %d %d\n",max,e,f);
if(n!=0) printf("\n"); //不是最后一个 就输出换行符
}
}
这个是参考另一个博主 不是自己写的!!!!