1003:max sum

    技术2024-10-26  21

    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

     

    代码:

     

    #include<cstdio>int a[1000];

    int  main(){ int cas; scanf("%d",&cas); for(int i=1;i<=cas;i++) {  printf("Case %d:/n",i);  int n,beg=1,end=1,temp=1;  scanf("%d",&n);  int sum=0,max=-9999;  for(int j=1;j<=n;j++)  {   int num;   scanf("%d",&num);   sum+=num;   if(sum>max)   {    max=sum;    end=j;    beg=temp;   }   if(sum<0)   {    sum=0;    temp=j+1;   }  }  printf("%d %d %d/n",max,beg,end);  if(i!=cas)   printf("/n"); } return 0;}

     

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