hdu1003 MaxSum (DP)

hdu1003 MaxSum (DP)

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 180281    Accepted Submission(s): 42103

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

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

Author

Ignatius.L

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#include #define num 100000int main(){ int n,t,i,j,flag,a[num],star,q,g; long sum,maxsum; scanf("%d",&t); for(i=1;i<=t;i++) { scanf("%d",&n); for(j=0;jmaxsum) { maxsum=sum; flag=q; star=j; } } printf("Case %d:\n",i); printf("%ld %d %d",maxsum,star+1,flag+1); if(i==t) printf("\n"); else printf("\n\n"); } return 0;}

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