≡ r?2 (mod n) as ≡ (xr)?2 (mod n)
SIPHASH (headerHash, 2nonce) mod 31.
Step 3: Alice calculates the common value x . . . . . . . . . . . . . . . . . . . . . . . . Bob calculates the common value of y , g , b mod p , where mod is the mode operator.
Each element of group L1 corresponds to two elements of group L0, i.e. any y of group L1, and group L0 has two x and (-x) mod F, which meets the requirements of x?2 mod F?
Since Xi and n intersert, it is available: a( n) - 1 ≡ 0 mod n , a (n) ≡ 1 mod n, the theotic evidence.
g?(x?y)(mod p)?g?g?(y?x) (mod p)?g?xy (mod p) ?key for encryption.
In this way, both parties know that the key is g?(xy) mod n ?(g?y mod n)?x mod n?(g?x mod n)?y mod n.
zerop (mod tetris-n-shapes interval))
The reason is simple, because N is n1... The product of nk, so Y mod n1 x0, and we know that x0 meets the equation system (1), so x0 mod n1 is a1, so there is Y mod.
case use_cache (Mod, LServer) of.
Step 3: Alice calculates the common value x s g s a mod p. Bob calculates the common value of y , g , b mod p , wheremod is the mode operator.
go build -mod-vendor -o chain main.go.
There is a theor theority, 10 m mod 1_000_000_007_u64 , 10 , (m mod 1_000_000_006_u64) mod 1_000_000_007_u64, which can greatly speed up the calculation process and calculate the results in 25 seconds.