HDU--1003Max Sum【水题】

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158268    Accepted Submission(s): 37019

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

 

#include<cstdio>
#include<cstring>

int a[100005];

int main ()
{
    int T,con=1;
    scanf("%d",&T);
    while(T--){
        int st,ed,x,y,sum,newsum;
        st=ed=x=y=1;
        int n,i;
        scanf("%d",&n);
        for(i=1;i<=n;++i)
            scanf("%d",&a[i]);
        sum=newsum=a[1];
        for(i=2;i<=n;++i){
            if(sum>=0){
            sum+=a[i]; y=i;
            }
            else{
                sum=a[i];x=y=i;
            }
            if(sum>newsum){
                newsum=sum;st=x;ed=y;
            }
        }
        printf("Case %d:\n",con++);
        printf("%d %d %d\n",newsum,st,ed);
        if(T) printf("\n");
    }
    return 0;
}