超时处理 时间复杂度问题杭电1003



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
   
   
   
   

 

Author

Ignatius.L

 

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 题意：求连续子序列的最大和并且输出最大值与起始于终了的位置。

思路：dp嘛，连续求和就好结果超时了，发现给的是100000，两个for循环就报了，看了别人的处理办法。。。。。

看代码说好了：

#include<stdio.h>
#include<string.h>
int a[100005];
int main()
{
   int i,n,m,j,h,start,end,b;
   int sum;
   scanf("%d",&n);
   for(i=1;i<=n;i++)
   {
      sum=-100000;//最大值
      scanf("%d",&m);
      b=0;//求和
      h=0;
      start=0;
      end=0;
      for(j=0;j<m;j++)
      {
          scanf("%d",&a[j]);
          b=b+a[j];
         if(b>sum)
         {
             sum=b;
             start=h;
             end=j;
         }
         if(b<0)//这一步很重要，当计算到这一步发现小于0，就从下一个位置重新连续相加。。。
         {
             h=j+1;
             b=0;
         }
      }
      printf("Case %d:\n%d %d %d\n",i,sum,start+1,end+1);
      if(i<n)
      {
          printf("\n");
      }
   }
 return 0;
}//

看见水题一激动就错，以后敲代码前自己先仔细想想好了，想清楚再做。。。。。