题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158421 Accepted Submission(s):
37055
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.
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).
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.
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
题目大意：给一串数字，然后要求出和最大的连续子序列。并且题目要求这个和最大的连续子序列输出起始位置和终止位置。
注意:初始化问题，还有空格的这个格式问题！
详见代码。
#include <iostream>
#include <cstdio> using namespace std; struct node
{
int s,e,smax;
} sum[]; int main ()
{
int t,n,Max,k,j;
int num[];
while (~scanf("%d",&t))
{
int flag=;
while (t--)
{ scanf("%d",&n);
for (int i=; i<n; i++)
{
scanf("%d",&num[i]);
}
sum[].smax=num[];
sum[].s=sum[].e=;
k=,j=;
Max=sum[].smax;
for (int i=; i<n; i++)
{
if (num[i]>sum[i-].smax+num[i])
{
sum[i].smax=num[i];
sum[i].s=i;
sum[i].e=i;
}
else
{
sum[i].smax=sum[i-].smax+num[i];
sum[i].s=sum[i-].s;
sum[i].e=i;
}
if (Max<sum[i].smax)
{
Max=sum[i].smax;
k=sum[i].s;
j=sum[i].e;
}
}
printf ("Case %d:\n",flag++);
printf ("%d %d %d\n",Max,k+,j+);
if (t)
printf ("\n");
}
}
return ;
}