A

Total Submission(s) : 210   Accepted Submission(s) : 40

Problem Description

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.<br>

 

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).<br>

 

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.<br>

 

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

 

题目大意：求个子数列让这个数列的和最大；

这个数列是连续的，并且要把子列的位置记录下来；

思路：

因为他要有起始位置和结束位置所以我就从这里入手定义了一个二维数组来存起始位置和和

最后比较去最大的和输出和其实位置和结束位置

代码：

#include<iostream>

#include<string.h>

using namespace std;

int iarr[100001];

int  dp[100001][2];

int n;

int ans = -100000000,i,p;

void dps()

{

ans = -100000000;

 dp[0][0] = iarr[0];

 dp[0][1] = 0;

 for (i = 1; i < n; i++)

 {

  if (dp[i - 1][0] +  iarr[i] >=  iarr[i])

  {

   dp[i][0] = dp[i - 1][0] +  iarr[i];

   dp[i][1] = dp[i - 1][1];

  }

  else

  {

   dp[i][0] =  iarr[i];

   dp[i][1] = i;

  }}

 for (i = 0; i < n; i++)

  if (dp[i][0] > ans)

  {

   ans = dp[i][0];

   p = i;

  }

}

int main()

{

    int t;

    cin>>t;

    for(int s=1;s<=t;s++)

    {

        memset(iarr,0,sizeof(iarr));

        memset(dp,0,sizeof(dp));

        cin>>n;

    for(int j=0;j<n;j++)

        cin>>iarr[j];

        dps();

         cout << "Case " <<s<< ":" <<endl;

        if (s != t)

            cout << ans << " " << (dp[p][1] + 1) << " " << (p + 1) << endl <<endl;

        else

            cout << ans << " " << (dp[p][1] + 1) << " " << (p + 1) << endl;

    }

return 0;

}