HDU 1003 Max Sum

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

 //话说这题当时我是看着代码，带入数据才明白的，好几个动态规划都是这样才明白的

#include <iostream>//简单的动态规划
#include <algorithm>
#include <cstdio>
#include <string.h>
#include <queue>
#include <stdlib.h>
using namespace std;
int main()
{   //freopen("in.txt","r",stdin);
    int t,n,a,max,b,i,f,l,temp,flag=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&b);
        max=b;temp=f=l=1;
        if(flag)
         printf("\n");
        flag++;
        for(i=2;i<=n;i++)
        {
            scanf("%d",&a);
            if(b<0)     //关键的代码
              {b=a;temp=i;}
            else
              b+=a;
            if(b>max)//关键的代码
             {
                max=b;
                f=temp;
                l=i;
             }
        }
      printf("Case %d:\n%d %d %d\n",flag,max,f,l);
    }
    return 0;
}
                                                                  ------江财小子