HDU 1003

Max Sum

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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 242877 Accepted Submission(s): 57348

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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.

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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).

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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.

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Sample Input

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

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Sample Output

Case 1:
14 1 4

Case 2:
7 1 6

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最大值可以是负的。

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#include<iostream>#include<stdio.h>using namespace std;int main(){    int t,n,a,l,r,left,mx,sum;    cin>>t;    int cas=0;    while(t--)    {        cas++;        scanf("%d",&n);        sum=0;        mx=-1010;        l=r=left=1;        for(int i=1;i<=n;i++)        {            scanf("%d",&a);            sum+=a;            if(sum>mx)            {                mx=sum;                r=i;                left=l;            }            if(sum<0)                sum=0,l=i+1;        }        printf("Case %d:\n",cas);        printf("%d %d %d\n",mx,left,r);        if(t)            printf("\n");    }    return 0;}