hdoj 1003 Max Sum

Max Sum

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 11   Accepted Submission(s) : 5

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

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

#include<stdio.h> #include<string.h> #define MAX 100000+10 #define INF 0x3f3f3f int main() {     int n,m,j,i,k1,k2,t,sum;     int x,y,sum1;     int s[MAX];     scanf("%d",&t);     m=1;     while(t--)     {         memset(s,0,sizeof(s));         scanf("%d",&n);         for(i=0;i<n;i++)         {             scanf("%d",&s[i]);         }         x=y=k1=k2=0;         sum=s[0];         sum1=sum;         for(i=1;i<n;i++)         {             if(sum>=0)             {                 y=i;                 sum+=s[i];             }             else             {                 x=i;                 y=i;                 sum=s[i];               }             if(sum>sum1)             {                 sum1=sum;                 k1=x;                 k2=y;             }         }         printf("Case %d:\n",m++);         printf("%d %d %d\n",sum1,k1+1,k2+1);         if(t!=0)         printf("\n");     }     return 0; }