Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 245162 Accepted Submission(s): 57896
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
题目解析：
这是一个简单的DP题，类似于最大上升子序列问题，我们只要可以求出最大值就可以输出开始和结尾了
代码：
#include <iostream>
#include<cstdio>
using namespace std;
int main()
{
int j,i,k,n,m,t;
int a[100002];
scanf("%d",&t);
for (j=1;j<=t;j++)
{
scanf("%d",&n);
for (i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
int sum=0,maxsum=-1001,first =0, last = 0, temp = 1;
for (i=0;i<n;i++)
{
sum += a[i];
if (sum > maxsum)
{
maxsum = sum;first = temp;last = i+1;
}
if (sum < 0)
{
sum = 0;temp = i+2;
}
}
printf("Case %d:\n%d %d %d\n",j,maxsum,first,last);
if (j!=t)
{
printf("\n");
}
}
return 0;
}
