Uva11440-help tomisu (number theory)

Test instructions: Given N and M, ask [2,n!] , the number of all quality factors is greater than the number of M

Ideas: ? ( m ! < /span> is less than m! and with the number of m! coprime, and the number of m! coprime, his quality factor certainly does not contain 1-m, so is to meet the conditions. Then for this problem, it is required n!, not with the number of m! coprime, answer modulo 100000007

So first look at a proof:
To find the number of coprime and N in kn, the answer is k? (n) 。
? ( n < /span> Represents the number of 1-n and N coprime, then consider [n + 1, 2n], [2n + 1, 3n] ... Each of the numbers in each interval is equal to 1-n in the number plus kn, for the original and n is not coprime number, plus n will still have a qualitative factor repetition, so still not, then for the original coprime number X,GCD (x, N) = 1, then the GCD (x + kn, n) = 1, is still coprime, So every n interval is the same as the number of n coprime, so the answer k? (n)

So for this question, the answer becomes n! / m ! ? (m! ) ), then the question remains how to ask? (m! ) )。

Known? (n) To find the lawN?(1?1/ p 1 )?(1?1/ p 2 )....(1?1/ p n )(P is the mass factor of N), so m! , the molecule is m! , the denominator is the 1-m of all prime numbers. ( 1 1 / p multiply product

The answer here is to ask, m! Eliminate, getN!/∏(1?1/ p i )mod 1000000007, first preprocessing those tables, each time to calculate can

#include <cstdio> #include <cstring>const long long N = 10000005;const long Long mod = 100000007;long long ispri[ N],fac[n],phi[n];long long EXGCD (Long long A,long long B,long long &x,long long &y) {    if (!b) {X=1;y=0;return A ; A    long long d=exgcd (b,a%b,y,x);    Y-= a/b*x;    return D;} Long Long A,long (long long) {Long long n    x, y;    EXGCD (a,n,x,y);    Return (X+n)%n;} void Get_table () {    fac[0]=fac[1]=1; phi[0]=phi[1]=1;    For (long long i=2;i<n;i++) {        fac[i]= (fac[i-1]*i)%mod;        if (Ispri[i]) {            phi[i]=phi[i-1];            Continue;        }        phi[i]=phi[i-1]* (i-1)%MOD*INV (i,mod)%mod;        For (long long j=i*i;j<n;j+=i)            ispri[j]=1;}    } int N,m;int Main () {    get_table ();    while (~SCANF ("%d%d", &n,&m) &&n) {        printf ("%lld\n", ((fac[n]*phi[m]-1)%mod+mod)%mod);    }    return 0;}

Recently brush problems feel good, light attention to the problem of the feeling is particularly busy, summary has not written. The recent courses have also fallen a lot ... Brush up the problem first!

Uva11440-help tomisu (number theory)