hdu1003

                                                           Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52833    Accepted Submission(s): 11726

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output
Case 1:14 1 4Case 2:7 1 6

 

 

#include<stdio.h>  int main()  {      int n;      int a[100005];      int t;      int k=1;      scanf("%d",&t);      while(t--)      {          scanf("%d",&n);          for(int i=0; i<n; i++)              scanf("%d",&a[i]);          int sum,max_sum,begin,end,pos;          sum=max_sum=a[0];        //初始化          begin=end=pos=0;                   for(int j=1; j<n; j++)   //循环n-1次，把剩下的n-1个数全部处理，因为第一个数已经放到sum里了          {              sum+=a[j];           //把a[j]加到sum里              if(sum<a[j])         //没加a[j]之前的sum < 0,更新sum和pos              {                  pos=j;                  sum=a[j];              }              if(sum>max_sum)      //更新max_sum，begin和end              {                  max_sum=sum;                  end=j;                  begin=pos;              }          }          printf("Case %d:\n",k++);          printf("%d %d %d\n",max_sum,begin+1,end+1);    //输出结果          if(t)  printf("\n");      }      return 0;  }  

贪心：开一个变量max_sum来记录连续子序列和的最大值，用begin记录max_sum对应序列的起点位置，用end记录max_sum对应序列的终点位置，用sum记录当前序列的连续和，pos记录当前sum的起点位置。每次都将依次将一个数a[j]加到sum里去,如果sum < a[j]， 那就代表不加a[j]之前的sum < 0，这样的话，我们就把前一段的舍掉，更新pos = j，sum = a[j]，让新的一段子序列从位置j开始，如果这个时候的 sum > max_sum,那么我们就更新max_sum,同时也应该更新max_sum所对应的起点和终点位置。重复上述操作，直到把所有的序列全部加到sum。这时候的max_sum，begin，end即为所求。