Max Sum

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

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;  
}