<hdoj1003>Max Sum

Description

Given a sequence a[1],a[2],a[3]......a
, 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
这道题本是dp练习题  我感觉还是贪心好做一点  注意全是负数的情况
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
#define  INF  1<<30
int  main()
{
int n;
int t;
scanf("%d",&t);
int cut=1;
while(t--)
{
scanf("%d",&n);
int xu[100000];
int qi=1;
int wei=1;
int max=0;
int sum=0;
int flog=0;
int min=-INF,qi1,wei1;
for(int i=1;i<=n;i++)
{
scanf("%d",&xu[i]);
if(xu[i]>0)
{
flog=1;
}
}
int r=1;
for( int i=1;i<=n;i++)
{
sum+=xu[i];
if(xu[i]>min)
{
min =xu[i];
qi1=i;
wei1=i;
}
if(sum<0)
{
sum=0;
r=i+1;
}
if(sum>=max)
{
max=sum;
qi=r;
wei=i;
}

}
printf("Case %d:\n",cut++);
if(flog==0)
{
printf("%d %d %d\n",min,qi1,wei1);
}
else
{
printf("%d %d %d\n",max,qi,wei);
}
if(t!=0)
printf("\n");
}
return 0;
}