HDU 1003 Max Sum 最大连续上升和

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 139206    Accepted Submission(s): 32328

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4Case 2:7 1 6

/*HDU 1003 最长上升序列和 用len1 len2表示起始结束的位置 */#include <iostream>using namespace std;int a[100001];int main(){    int n,m,i,max,sum,len1,len2,k,j;    cin>>m;    k=1;    while (m--)    {        scanf("%d",&n);        max=-1001;        sum=0;        len1=1;        len2=1;        for(i=0;i<n;i++)        {            scanf("%d",&a[i]);            if(sum<0)                 sum=a[i];//正的就往上加            else sum+=a[i];                        if(sum>max)                {                    max=sum;                     len2=i+1;                    }        }                for(j=0;j<n;j++)            {            sum=0;            for(i=j;i<len2;i++)                sum+=a[i];            if(sum==max)                {                    len1=j+1;                    break;                }        }                  printf("Case %d:\n",k++);        printf("%d %d %d\n",max,len1,len2);        if(m!=0) printf("\n");    }    return 0;}