hdu 1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 255288    Accepted Submission(s): 60663

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

 

 

题解：最简单的尺取法     一次遍历 

一路加上去   如果累加的值sum    sum小于0了    就记录开头的节点

如果sum大于了记录最大值的ans   就更新ans的值  和开头结尾

//尺取法 

#include<stdio.h>
#include<string.h>
int que[100002];
main()
{
  int t,flag;
  scanf("%d",&t);
  flag=t;
  while(t--)
  {
    memset(que,0,sizeof(que));
    int n,i,temp=0;
    int fir=1,end=1,sum=-1,max=-0x7fffffff;/*0x7fffffff这是2进制的int最大*/
    scanf("%d",&n);
    for(i=1;i<=n;i++)
      scanf("%d",&que[i]);
    for(i=1;i<=n;i++)
    {
      if(sum<0)
      {
        temp=i;
        sum=que[i];
        
      }
      else
      {
          sum+=que[i];
          
      }
      if(sum>max)
        {
             max=sum;
            fir=temp;
            end=i; 
        }
         
    }
    
    printf("Case %d:
",flag-t);
    printf("%d %d %d
",max,fir,end);
    if(t)
        printf("
");
  }
  return 0;
}