思维(Python)
小红的平滑值插值
https://www.nowcoder.com/practice/97f6b71b1df745cd9926eb861e2d89df
思路:整体思路比较容易想到:当两数间的差值时,用公差为
的等差数列去从小到大的插入,结果就是
,然后不断累加即可。但是有个特判不好注意到,整体思路可以处理差值中有k的情况,此时会正确输出0;但当所有差值都< k时,此时应该输出1,因为我们要额外构造一个数,使得差值为k。所以说再额外引入一个flag数组,作为特判的标记
代码:
import sys
input = lambda: sys.stdin.readline().strip()
import math
inf = 10 ** 18
def I():
return input()
def II():
return int(input())
def MII():
return map(int, input().split())
def GMI():
return map(lambda x: int(x) - 1, input().split())
def LI():
return input().split()
def LII():
return list(map(int, input().split()))
def LFI():
return list(map(float, input().split()))
fmax = lambda x, y: x if x > y else y
fmin = lambda x, y: x if x < y else y
isqrt = lambda x: int(math.sqrt(x))
'''
'''
def solve():
n, k = MII()
a = LII()
cnt = 0
flag = [False, False]
for i in range(1, n):
if abs(a[i] - a[i - 1]) > k:
cnt += (abs(a[i] - a[i - 1]) - 1) // k
flag[0] = True
if abs(a[i] - a[i - 1]) == k:
flag[1] = True
if flag[0]:
print(cnt)
elif flag[1]:
print(0)
else:
print(1)
t = 1
# t = II()
for _ in range(t):
solve()
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