谐距下标对_

Python3

三文鱼在发呆

from collections import Counter

n = int(input())
numbers = list(map(int, input().split()))

b = [numbers[i] - (i+1) for i in range(n)]
counter = Counter(b)
count = 0
for freq in counter.values():
    if freq >1:
        count += freq*(freq - 1) //2
print(count)

发表于 2025-10-25 23:44:22 回复(0)

Python3

xin202309

import sys

for line in sys.stdin:

a = line.split()

an=list(map(int,a))

vdict={}

counts=0

for i in range(len(an)):

v=an[i]-i

if v not in vdict:

vdict[v]=1

else:

counts+=vdict[v]

vdict[v]+=1

print(counts)

发表于 2025-09-14 21:20:16 回复(0)

Python3

忧伤的双曲线

n = int(input())

a = list(map(int,input().split()))

for i in range(n):

a[i] = a[i]-i

dic = {}

count = 0

for i in range(n):

if a[i] not in dic.keys():

dic[a[i]] = 1

else:

count += dic[a[i]]

dic[a[i]] += 1

print(count)

发表于 2025-08-26 15:23:32 回复(0)

Python3

牛客847294653号

n = int(input())

a = list(map(int, input().strip().split()))

# 计算每个元素的"值减下标"

diff = [a[i] - i for i in range(n)]

# 统计相同差值的出现次数

count_map = {}

for val in diff:

count_map[val] = count_map.get(val, 0) + 1

# 计算谐距下标对数量

pairs = 0

for count in count_map.values():

# 如果有k个元素有相同的"值减下标"，则可以形成C(k,2)对

if count > 1:

pairs += count * (count - 1) // 2

print(pairs)

发表于 2025-08-22 10:33:14 回复(0)

谐距下标对

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