hdoj maxsum 1003



/*Max Sum
Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 5   Accepted Submission(s) : 1
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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.
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*/

<span style="font-size:18px;">#include<stdio.h>int main(){    int i,ca=1,t,s,e,n,x,now,before,max;    scanf("%d",&t);    while(t--)    {       scanf("%d",&n);       for(i=1;i<=n;i++)       {         scanf("%d",&now);         if(i==1)//初始化          {            max=before=now;//max保留之前算出来的最大和，before存储目前在读入数据前保留的和，   //now保留读入数据              x=s=e=1;//x用来暂时存储before保留的和的起始位置，当before>max时将赋在s位置，//s，e保留最大和的start和end位置          }         else {             if(now>now+before)//如果之前存储的和加上现在的数据比现在的数据小，    //就把存储的和换成现在的数据，反之就说明数据在递增，可以直接加上              {                before=now;                x=i;//预存的位置要重置              }                    else before+=now;              }         if(before>max)//跟之前算出来的最大和进行比较，如果大于，位置和数据就要重置            max=before,s=x,e=i;       }       printf("Case %d:\n%d %d %d\n",ca++,max,s,e);       if(t)printf("\n");     }    return 0;}</span>