鉴于n
的数量。如何找到所有这些不同元组的数量?
(i, j, k)
(i*j)%k == 0,
在哪里1<=i<=n, 1<=j<=n, 1<=k<=n
在O(n^2)
或更好。
示例代码:
vector<int>freq(n*n+1); //that's just array of n*n+1 zeroes
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
freq[i*j]++;
int cnt = 0;
for(int k=1; k<=n; k++)
for(int j=0; j<=n*n; j+=k) //note these loops look like n^3 at first glance yet it's something like n^2 log n (check harmonic number if you're curious why)
cnt+=freq[j];