题解 | 谐距下标对

#i-j=a[i]-a[j]即diff=a[i]-i=a[j]-j
#用双循环会导致算法超时
n = int(input())
an = list(map(int,input().split()))
ans = 0#答案
counts = {}#a[i]-i的个数
for i in range(n):#
    diff = an[i]-i
    ans += counts.get(diff,0)#统计当前能匹配的个数
    counts[diff] = counts.get(diff,0)+1#当前能匹配的个数加一
print(ans)#输出结果