Basic algorithm POJ 1845 spoj divsum in Number Theory

Extended Euclidean inverse element (as long as a and B are mutually unique, there is a inverse element)
[Cpp]
# Include <cstdio>
# Include <cstring>
# Include <cstdlib>
# Include <cmath>
# Define mod 9973.
Typedef long ll;
Void ExGcd (int a, int B, int & x, int & y ){
If (B = 0 ){
X = 1;
Y = 0;
Return;
}
ExGcd (B, a % B, x, y );
Int t = x;
X = y;
Y = t-a/B * y;
}
Int main (){
Int t, T, rev, x;
Int a, B;
Scanf ("% d", & T );
For (t = 1; t <= T; t ++ ){
Scanf ("% d", & a, & B );
ExGcd (B, mod, rev, x );
Printf ("% d \ n", (rev * a) % mod + mod) % mod );
}
}

 

O (n) base number table

[Cpp]
Int pn = 0, prime [MAXN], factor [MAXN]; // The minimum approximate number of elements whose factor [I] Is I
Void get_prime (int n ){
Int I, j;
For (I = 1; I <= n; I ++)
Factor [I] = I;
For (I = 2; I <= n; I ++ ){
If (I = factor [I]) prime [pn ++] = I;
For (j = 0; j <pn & prime [j] * I <= n; j ++ ){
Factor [I * prime [j] = prime [j];
If (I % prime [j] = 0)
Break;
}
}
}

POJ 1845 binary (1 + x ^ 2... + x ^ n) summation
[Cpp]
# Include <cstdio>
# Include <cstring>
# Include <cstdlib>
# Include <cmath>
# Define mod 9901.
Typedef long ll;
Const int MAXN = 10000;
Int a, B, k;
Int p [MAXN], n [MAXN];
Ll quick (ll num, ll mi ){
Ll ans = 1;
While (mi ){
If (mi % 2) ans = (ans * num) % mod;
Num = (num * num) % mod;
Mi/= 2;
}
Return ans % mod;
}
Ll sum (ll num, ll mi ){
If (mi = 0) return 1;
If (mi % 2)
Return (sum (num, mi/2) * (1 + quick (num, mi/2 + 1) % mod;
Return (sum (num, mi/2-1) * (1 + quick (num, mi/2 + 1) + quick (num, mi/2) % mod;
}
Void factor (int ){
Int I;
K = 0;
For (I = 2; I * I <= a;) {// complexity decomposition INTEGER OF sqrt (n)
If (a % I = 0 ){
P [k] = I;
N [k] = 0;
While (a % I = 0 ){
N [k] ++;
A/= I;
}
K ++;
}
If (I> 2) I + = 2;
Else I ++;
}
If (a> 1 ){
P [k] =;
N [k] = 1;
K ++;
}
}
Int main (){
Int I;
Scanf ("% d", & a, & B );
Factor ();
Int ans = 1;
For (I = 0; I <k; I ++)
Ans = (ans * (sum (p [I], n [I] * B) % mod;
Printf ("% d \ n", ans );
}
POJ 1845 reverse RMB Method
[Cpp]
# Include <cstdio>
# Include <cstring>
# Include <cstdlib>
# Include <cmath>
# Define mod 9901.
Typedef long ll;
Const int MAXN = 10000;
Int a, B, k;
Int p [MAXN], n [MAXN];
 
Void ExGcd (int a, int B, int & x, int & y ){
If (B = 0 ){
X = 1;
Y = 0;
Return;
}
ExGcd (B, a % B, x, y );
Int t = x;
X = y;
Y = t-a/B * y;
}
Int reverse (int num ){
Int rev, x;
ExGcd (num, mod, rev, x );
Return rev;
}
Ll quick (ll num, ll mi ){
Ll ans = 1;
While (mi ){
If (mi % 2) ans = (ans * num) % mod;
Num = (num * num) % mod;
Mi/= 2;
}
Return ans % mod;
}
Ll sum (ll num, ll mi ){
If (num % mod = 1) // note that this situation cannot be reversed because the num-1 and mod are not mutually exclusive
Return (mi + 1) % mod;
Return (quick (num, mi + 1)-1 + mod) % mod) * reverse (num-1) % mod + mod) % mod;
}
Void factor (int ){
Int I;
K = 0;
For (I = 2; I * I <= a;) {// complexity decomposition INTEGER OF sqrt (n)
If (a % I = 0 ){
P [k] = I;
N [k] = 0;
While (a % I = 0 ){
N [k] ++;
A/= I;
}
K ++;
}
If (I> 2) I + = 2;
Else I ++;
}
If (a> 1 ){
P [k] =;
N [k] = 1;
K ++;
}
}
Int main (){
Int I;
Scanf ("% d", & a, & B );
Factor ();
Int ans = 1;
For (I = 0; I <k; I ++)
Ans = (ans * (sum (p [I], n [I] * B) % mod + mod) % mod;
Printf ("% d \ n", ans );
}

SPOJ DIVSUM

[Cpp]
# Include <cstdio>
# Include <cstring>
# Include <cstdlib>
# Include <cmath>
 
Const int MAXN = 1501000, N = 500010;
 
Int pn = 0, prime [MAXN], factor [MAXN];
Void get_prime (int n ){
Int I, j;
For (I = 1; I <= n; I ++)
Factor [I] = I;
For (I = 2; I <= n; I ++ ){
If (I = factor [I]) prime [pn ++] = I;
For (j = 0; j <pn & prime [j] * I <= n; j ++ ){
Factor [I * prime [j] = prime [j];
If (I % prime [j] = 0)
Break;
}
}
}
 
Int main ()
{
Int cases;
Get_prime (N );
Scanf ("% d", & cases );
While (cases --)
{
Int n, tmpn;
Long ans = 1;
Scanf ("% d", & n );
Tmpn = n;
While (tmpn! = Factor [tmpn])
{
Long fac = factor [tmpn], mtp = fac;
While (tmpn % fac = 0)
{
Mtp * = fac;
Tmpn/= fac;
}
Ans * = (1-mtp)/(1-fac );
}
If (tmpn> 1)
Ans * = (1 + tmpn );
Ans-= n;
Printf ("% lld \ n", ans );
}
 
Return 0;
}

SPOJ DIVSUM ANOTHER METHOD
[Cpp]
# Include <cstdio>
Using namespace std;
Int sum [500100];
Int main ()
{
Int T, t, I, j, n;
Scanf ("% d", & T );
For (I = 1; I <= 500000; I ++ ){
For (j = 2; I * j <= 500000; j ++) {// If j starts from 1, it will calculate itself
Sum [I * j] + = I;
}
}
For (t = 1; t <= T; t ++ ){
Scanf ("% d", & n );
Printf ("% d \ n", sum [n]);
}
}