Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 177961    Accepted Submission(s): 41505


[align=left]Problem Description[/align]
Given a sequence a[1],a[2],a[3]......a
, 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.

 

[align=left]Input[/align]
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).

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

 

[align=left]Sample Output[/align]

Case 1:
14 1 4

Case 2:
7 1 6 

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C语言程序代码

#include<stdio.h>

#include<string.h>

int a[100010];

int main(){

 int t,cnt=1;

 scanf("%d",&t);

 while(t--)

 {

  int st,ed,x,y,sum,sumn;

  st=ed=x=y=1;

  int n,i;

  scanf("%d",&n);

  for(i=1;i<=n;i++)

   scanf("%d",&a[i]);

  sum=0;sumn=a[1];

  for(i=1;i<=n;i++)

  {

   sum+=a[i];

   if(sum>sumn)//通过比较更新最大和，与起点和终点。

   {

    sumn=sum;

    st=x;ed=i;

   }

   if(sum<0)

   {

    x=i+1;

    sum=0;

   }

  }

  printf("Case %d:\n",cnt++);

  printf("%d %d %d\n",sumn,st,ed);

  if(t)

   printf("\n");

 }

 return 0;

}