POJ 1845 Sumdiv (arithmetic basic theorem to find a number factor and)

To find all the factors of a number and to use the basic theorem of arithmetic, here are two important applications:
(1) A positive integer greater than 1 N, if its standard decomposition is: n= (P1^A1) * (P2^A2) ... (Pn^an)
Then it's a number of numbers (1+A1) (1+A2) ..... (1+an).
(2) The sum of its all positive factors is d (N) = (1+p1+...p1^an) (1+P2+...P2^A2) ... (1+pn+...+pn^an)
And the method of finding a number is analogous to a number of numbers.
It is possible to first make a list of all the primes within sqrt (n) (and certainly not the table), since only one of the prime factors of n is greater than sqrt (n) (as evidenced in the previous topic),
So it can be processed at the end. For each item is a geometric series, summation is easy.


The following is an example of POJ 1845 Sumdiv.
The n^m%p,p is the prime number 9901.
1) because N and M are large, you can handle all the factor of N, then ^m the exponent of each factor *m. This problem can not be preprocessed all prime numbers, directly decomposition factor, for the time complexity of the problem is acceptable;
2) Use the fast power operation.

3) When seeking geometric series and modulus of two methods: 1. Summation formula with geometric series:s=a1* (q^n-1)/(q-1), to use the inverse of (q-1) modulo 9901, can be obtained by Euler theorem or extended Euclidean, but because the mod is a prime number, So the inverse of a is a^ (mod-2)%mod, with the fast power to seek. 2. Recursive form of dichotomy

for inverse complement: The Fermat theorem (Fermat theory) is an important theorem in number theory, with the following content: If P is prime, and gcd (a,p) = 1, then a (p-1) ≡1 (mod p)that is: If A is an integer, p is a prime number, and A,p coprime (that is, there is only one convention of 1), then the remainder of a (p-1) divided by P is constant equal to 1.

Inverse element method for solving

#include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include <cmath > #define MAXN 10000#define ll long longusing namespace Std;const ll mod=9901;ll prime[maxn],num,flag[maxn+5];ll Pow_mo    D (ll a,ll b,ll p) {//fast power ll ret=1;        while (b) {if (b&1) ret= (ret*a)%p;        A= (a*a)%p;    b>>=1; } return ret;}    void Get_prime () {//Sieve prime memset (flag,0,sizeof (flag));    num=0;        for (ll i=2;i<maxn;i++) {if (!flag[i]) prime[++num]=i;            for (ll j=1;j<=num&&i*prime[j]<maxn;j++) {flag[i*prime[j]]=1;        if (i%prime[j]==0) break;    }}}ll get_sum (ll k,ll N) {//calculation (1+k+k^2+...+k^n)%mod, geometric series and for (k^ (n+1)-1)/(K-1);    if (n==0) return 1;    ll Ret=1;    Ret=ret* (Pow_mod (K,n+1,mod)-1)%mod;       Ret=ret*pow_mod (k-1,mod-2,mod)%mod; (k-1) of the inverse, to note (k-1)%9901! = 0 return ret;}    int main () {ll n,m,k,a;    Get_prime (); while (~SCANF ("%lld%lld", &n,&m)) {ll ans= 1;     if (!n&&m) ans=0;        0^0=1, 0^1..=0;            The factor decomposition for (int i=1;i<=num&&prime[i]*prime[i]<=n;i++) {if (n<prime[i]) breaks;               if (n%prime[i]==0) {a=0;                    exponential while (n%prime[i]==0) {n/=prime[i];                a++;                } a*=m;                K=prime[i];            Ans=ans*get_sum (k,a)%mod; }}//If there is also a vegetarian factor greater than sqrt (n), its exponent must be 1*m=m if (n>1&& (n-1)%mod==0) ans=ans* (Pow_mod (n,m+1,mod* (n   -1))/(n-1))%mod;        if (n-1)%mod==0, cannot seek inverse meta-if (n>1) ans=ans*get_sum (n,m)%mod;                printf ("%lld\n", ans); I didn't add mod wa several times at first, because this method ans may be negative} return 0;}

The solution of the dichotomy method

#include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include <cmath > #define MAXN 10000#define ll long longusing namespace Std;const ll mod=9901;ll prime[maxn],num,flag[maxn+5];ll Pow_mo    D (ll a,ll b,ll p) {//fast power ll ret=1;        while (b) {if (b&1) ret= (ret*a)%p;        A= (a*a)%p;    b>>=1; } return ret;}    void Get_prime () {//Sieve prime memset (flag,0,sizeof (flag));    num=0;        for (ll i=2;i<maxn;i++) {if (!flag[i]) prime[++num]=i;            for (ll j=1;j<=num&&i*prime[j]<maxn;j++accepted) {flag[i*prime[j]]=1;        if (i%prime[j]==0) break;    }}}ll get_sum (ll k,ll N) {//Can be computed with a recursive binary (1+k+k^2+...+k^n)%mod if (n==0) return 1;    if (n&1) return ((1+pow_mod (K,n/2+1,mod)) *get_sum (K,N/2))%mod; else Return ((1+pow_mod (K,n/2+1,mod)) *get_sum (k,n/2-1) +pow_mod (k,n/2,mod))%mod;}    int main () {ll n,m,k,a;    Get_prime (); while (~SCANF ("%lld%lld", &n,&m) {ll ans=1; if (!n&&m) ans=0;        0^0=1, 0^1..=0;            for (int i=1;i<=num&&prime[i]*prime[i]<=n;i++) {if (n<prime[i]) break;               if (n%prime[i]==0) {a=0;                    exponential while (n%prime[i]==0) {n/=prime[i];                a++;                } a*=m;                K=prime[i];            Ans=ans*get_sum (k,a)%mod;        }} if (n>1) ans=ans*get_sum (n,m)%mod;    printf ("%lld\n", ans); } return 0;}

POJ 1845 Sumdiv (arithmetic basic theorem to find a number factor and)