最大子序列求和模版

long max3(long a, long b, long c)
{
       if (a < b)
            a = b;
       if (a > c)
              return a;
       else
              return c;
}
long maxSumRec(const vector<int>& a, int left, int right)
{
       if (left == right)
       {
              if (a[left] > 0)
                     return a[left];
              else
                     return 0;
       }
       int center = (left + right) / 2;
       long maxLeftSum = maxSumRec(a, left, center);
       long maxRightSum = maxSumRec(a, center+1, right);
       //求出以左边对后一个数字结尾的序列最大值
       long maxLeftBorderSum = 0, leftBorderSum = 0;
       for (int i = center; i >= left; i--)
       {
              leftBorderSum += a[i];
              if (leftBorderSum > maxLeftBorderSum)
                     maxLeftBorderSum = leftBorderSum;
       }
       //求出以右边对后一个数字结尾的序列最大值
       long maxRightBorderSum = 0, rightBorderSum = 0;
       for (int j = center+1; j <= right; j++)
       {
              rightBorderSum += a[j];
              if (rightBorderSum > maxRightBorderSum)
                     maxRightBorderSum = rightBorderSum;
       } 

       return max3(maxLeftSum, maxRightSum,
              maxLeftBorderSum + maxRightBorderSum);
}
long maxSubSum3(const vector<int>& a)
{
       return maxSumRec(a, 0, a.size()-1);
}

分治法 O(NlogN)