合數的質因數分解 + 遞歸求等比數列前n項和

質因數分解

const int maxn=10000;
bool book[maxn+10];
int prim[maxn+10],pnum=0;
void getprim()
{
    memset(book,0,sizeof(book));
    pnum=0;
    book[1]=book[0]=1;
    for(int i=2;i<=maxn;++i)
    {
        if(!book[i])
        {
            prim[pnum++]=i;
        }
        for(int j=0;j<pnum && prim[j]<=maxn/i ;++j)
        {
            book[prim[j]*i]=1;
            if(i % prim[j]==0) break;
        }
    }
}

ll fact[105][2];
int fnum=0;
int getfact(ll x)
{
    fnum=0;
    ll temp=x;
    for(int i=0;prim[i]<=temp/prim[i];++i)
    {
        fact[fnum][1]=0;
        if(temp%prim[i]==0)
        {
            fact[fnum][0]=prim[i];
            while(temp % prim[i]==0)
            {
                fact[fnum][1]++;
                temp /=prim[i];
            }
            fnum++;
        }
    }
    if(temp!=1)
    {
        fact[fnum][0]=temp;
        fact[fnum++][1]=1;
    }
    return fnum;
}

等比數列求和（從1開始）

ll sum(ll p ,ll n,ll mod)
{
    if(p==0) return 0;
    if(n==0) return 1;
    if(n&1) return (  ( 1+powermod(p,n/2+1,mod) )%mod * sum(p,n/2,mod)%mod )%mod;
    else return ( (1+powermod(p,n/2+1,mod)) %mod *sum(p,n/2-1,mod)+powermod(p,n/2,mod)%mod)%mod;
}