Problem - 1003_Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 151122    Accepted Submission(s): 35288

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

 

Author

Ignatius.L

 

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#include<stdio.h>int main(){    int t,n,addend,sum,sP,maxSum,startPos,endPos,i,j;    while(scanf("%d",&t)!=EOF){        for(i=1;i<=t;i++){            scanf("%d",&n);            for(j=1,sum=-1,maxSum=-1001;j<=n;j++){                scanf("%d",&addend);                if(sum>=0){                    sum+=addend;                }                else{                    sum=addend;                    sP=j;                }                if(sum>maxSum){                    maxSum=sum;                    startPos=sP;                    endPos=j;                }            }            printf("Case %d:\n%d %d %d\n",i,maxSum,startPos,endPos);            if(i!=t){                printf("\n");            }        }    }    return 0;}