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为了解决这个问题,我们需要找到一个方法来选择零或多个问题,使得解决它们的总时间不超过给定的时间限制,并且尽可能长。考虑到问题的数据范围较大,普通的动态规划方法可能会超时,因此我们采用折半搜索和二分法来优化解决方案。
我们可以将问题分成两部分,分别计算前半部分和后半部分的最大时间,然后合并这两部分的结果。具体步骤如下:
这种方法通过分治和二分法优化了时间复杂度,适用于较大的数据范围。
def main(): import sys input = sys.stdin.read data = input().split() n = int(data[0]) t = int(data[1]) a = list(map(int, data[2:2+n])) if n == 0: print(0) return # Function to compute the maximum possible sum not exceeding t def compute_max(arr): max_sum = 0 current_sum = 0 for num in arr: if current_sum + num > t: break current_sum += num if current_sum > max_sum: max_sum = current_sum return max_sum # Split into two halves mid = n // 2 sum1 = [] current_sum = 0 for i in range(mid): current_sum += a[i] if current_sum > t: break sum1.append(current_sum) # After mid, start adding from mid+1 current_sum = 0 for i in range(mid, n): current_sum += a[i] if current_sum > t: break sum1.append(current_sum) # Now compute for the second half in reverse sum2 = [] current_sum = 0 for i in range(n-1, mid-1, -1): current_sum += a[i] if current_sum > t: break sum2.append(current_sum) # Now add from mid to n-1 current_sum = 0 for i in range(mid, n): current_sum += a[i] if current_sum > t: break sum2.append(current_sum) # Sort the sum1 and sum2 sum1.sort() sum2.sort() # Now find the maximum sum1[i] + sum2[j] <= t max_total = 0 for i in range(len(sum1)): target = t - sum1[i] # Find the largest j where sum2[j] <= target low = 0 high = len(sum2) - 1 best = -1 while low <= high: mid = (low + high) // 2 if sum2[mid] <= target: best = mid low = mid + 1 else: high = mid - 1 if best != -1: current_total = sum1[i] + sum2[best] if current_total > max_total: max_total = current_total print(max_total)if __name__ == "__main__": main()
compute_max来计算前半部分或后半部分的最大时间,不超过时间限制。这种方法高效地解决了问题,确保在较短的时间内找到最优解。
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