Lightoj 1054-efficient Pseudo Code

Topic Connection:

http://www.lightoj.com/volume_showproblem.php?problem=1054

Main topic:

Given n,m, what is the sum of all the factors that ask n^m?

Problem Solving Ideas:

Additional knowledge:

1: For a digital n=p1^t1+p2^t2+p3^t3+.........+pn^tn. The sum of the added sum of the n factor and the corresponding value equivalent to the subset of n All factors-(P1^0+P1^1+P1^2+......+P1^T1) * (P2^0+P2^1+......+P2^T2) *.....* (PN^0+PN^1+......+PN^TN) .

2: Geometric series summation formula:

3: Division Take-Over: (A/b)%p = = a% (b%p)/b%p;

a/b%p = = a*b^ (p-2)%p; (when P is prime) proof: There is a Fermat theorem: P is a prime number, (b,p) = 1, b^ (p-1)%p = = 1,a/b%p = a/b*1%p = a/b*b^ (p-1)% = a*b^ (p-2)%p.

1#include <bits/stdc++.h>2 using namespacestd;3 4 Const intMoD =1000000007;5 Const intMAXN =7000;6typedefLong LongLL;7 intISPRIME[MAXN], prime[maxn*Ten];8 LL sum, K;9 Ten voidIsPrime () One{//filter out the number of prime numbers needed A     intI, J; -      for(i=2, k=0; i<70000; i++) -         if(!Prime[i]) the         { -Isprime[k + +] =i; -              for(J=i; j<70000; j+=i) -PRIME[J] =1; +         } -        //printf ("%d\n", k); + } A  at ll Pow (ll X, ll y) -{//Fast Power seeking x^y -LL num =1; -      while(y) -     { -         if(Y%2) innum = (num * x)%MoD; -x = (x * x)%MoD; toY/=2; +     } -     returnnum; the } * ll solve (ll X, ll y) $ {Panax Notoginseng LL num; -num = (Pow (x, y)-1); thenum = (num * POW (x1, mod-2)) %MoD; +     return(num + MoD)%MoD; A } the  + intMain () - { $LL t, N, m, L =0; $ IsPrime (); -scanf ("%lld", &t); -      while(T--) the     { -scanf ("%lld%lld", &n, &m);Wuyi LL A, B, I; thei =0; -sum =1; Wu          while(I <k) -         { About             if(1==N) $                  Break; -A = b =0;//count the number of prime factors and factors -             if(n% isprime[i] = =0) -             { AA =Isprime[i]; +                  while(n% isprime[i] = =0) the                     { -B + +; $N/=Isprime[i]; the                     } thesum = (sum * Solve (A, b*m+1) ) %MoD; the             } thei + +; -              in         } the         if(n! =1) thesum = (sum * Solve (n, m+1)) %MoD; Aboutprintf ("Case %lld:%lld\n", ++l, sum); the     } the     return 0; the}

Lightoj 1054-efficient Pseudo Code