calculate 1+a^1+a^2+...+a^d mod m (using BigMod algorithm)

Правка en2, от iftekhar, 2020-05-20 13:08:21

How can I solve this using bigmod algorithm?

If the series is 1+a+a^2+a^3+a^4+a^5 then this will be equal to (1+a^2+a^4)+a(1+a^2+a^4). How can I apply bigmod here!

BigMod: this is for calculating the mod of a large number like 2^100 or 10^18.

ll bigmod(ll a, ll b, ll m) { if(b==0) return 1; if(b%2==1) { ll p1 = a%m; ll p2 = bigmod(a,b-1,m); return (p1*p2)%m; } if(b%2==0) { ll h = bigmod(a,b/2,m); return (h*h)%m; } }

Теги #number theory

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  Rev. Язык Кто Когда Δ Комментарий
en5 Английский iftekhar 2022-05-26 15:39:35 2 Tiny change: '\n**BigMod: **\n This ' -> '\n**BigMod :**\n This '
en4 Английский iftekhar 2022-05-26 15:38:24 53
en3 Английский iftekhar 2022-05-26 15:35:39 67
en2 Английский iftekhar 2020-05-20 13:08:21 346
en1 Английский iftekhar 2020-05-20 12:23:28 232 Initial revision (published)