43.3 100 0 0
咋这么难呢
贴一下第二题代码
题目:小红有个数组,数组相邻长度差值最多为1,并且元素都是正整数。现在小红知道数组长度为n,数组和为m,小红想知道所有符合条件数组中,p位置最大值是多少(起始位置为1) 输入三个整数n,m,p 1<=p<=n<=m<=10**9 思路:二分查找check判断,难点在于怎么快速算出整个数组的最小值,贪心思想,p位置为mid,然后逐渐减一,需要判断两个边界条件,如果减到最小值了还没到边界,就要补1.
n,m,p = map(int,input().split()) def check(mid): count = 0 if p + mid - 1 > n: count += (2 * mid - n + p ) * (n - p + 1) / 2 else: count += (mid + 1) * mid / 2 + (n - p - mid + 1) if p - mid + 1 <= 0: count += (mid-1 + mid - p + 1) * (p - 1) / 2 else: count += mid * (mid - 1) / 2 + (p - mid) return count<=m left,right = 1, m - n + 1 while left < right: mid = left + (right - left + 1)//2 if check(mid): left = mid else: right = mid - 1 print(left)