HDU 4746 mophues Möbius third bomb

Test instructions: 1<=x<=n,1<=y<=m, which causes gcd (x, y) =k,k to be less than or equal to the number of P

Example: 24=2*2*2*3,k=4

Solution: Set F[n] to GCD (A, b) =n logarithm

F[d] is the logarithm of D|GCD (A, B)

F[n]=sigema (Mu[i],f[i*n]):

F[1]=mu[1]*f[1]+mu[2]*f[1*2]+...+mu[n]*f[1*n]

F[2]=mu[2]*f[2]+mu[2]*f[2*2]+...+mu[n]*f[2*n]

......

Sum=f[1]+f[2]+...+f[n]=g[1]*f[1]+g[2]*f[2]+...+g[n]*f[n]

Enumerate each I, then the multiples of I J for G[j] provide the contribution of mu[j/i], i.e. g[j]+=mu[j/i]

Because the number of the g[j to be K is the result of several numbers, it is opened to a two-dimensional array g[j][p] to represent the contribution of J to the number of p.

Need to use the method of block acceleration, or it will explode

#include <stdio.h> #include <string.h> #define MIN (a) (a) < (b)? A):(B)) #define MAX (a) < (b) ( b):(a)) #define LL __int64const int maxn=500005;int num[maxn];int prime[maxn];int mu[maxn];int factor[maxn];int MBS[MAXN    ][20];void Mobius () {memset (num,0,sizeof (num));    int all=0;    Mu[1]=1;    factor[1]=0;            for (int i=2;i<maxn;i++) {if (!num[i]) {prime[all++]=i;            Mu[i]=-1;                 Factor[i]=1;            The record is a number of characters} for (int j=0;j<all&&i*prime[j]<maxn;j++) {num[i*prime[j]]=1;            factor[i*prime[j]]=factor[i]+1;            if (I%prime[j]) {mu[i*prime[j]]=-mu[i];                } else {mu[i*prime[j]]=0;            Break }}} return;}    void Inti () {memset (mbs,0,sizeof (MBS)); for (int i=1;i<maxn;i++) for (int j=i;j<maxn;j+=i) MBS[J][FACTOR[I]]+=MU[J/I]; Contribution of each J in Factor[i] factors

/* The following is for block acceleration summation *    /for (int i=1;i<maxn;i++) for        (int j=0;j<19;j++)            mbs[i][j]+=mbs[i-1][j];

    for (int. i=0;i<maxn;i++) for        (int j=1;j<19;j++)            mbs[i][j]+=mbs[i][j-1];    return;} int main () {    int t,n,m,p;    Mobius ();    Inti ();    ll sum;    while (scanf ("%d", &t)!=-1)    {while        (t--)        {            scanf ("%d%d%d", &n,&m,&p);            if (p>=19)                    //Because 2^19>500000, so more than 19 is all meet the requirements            {                printf ("%i64d\n", (ll) n*m);                Continue;            }            if (n>m)            {                int te=n;                n=m;                M=te;            }            sum=0;            for (int i=1,last;i<n;i=last+1)            {                last=min (n/(n/i), m/(m/i));          

Block acceleration, because [n/i][m/i] has a repeating part in the I sliding scale process, skipping these I but simplifying the calculation, is the complexity reduced to sqrt (n)                sum+= ((LL) (n/i) * (m/i) * (mbs[last][p]-mbs[i-1][p ]));            }            printf ("%i64d\n", sum);        }    }    return 0;}

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HDU 4746 mophues Möbius third bomb