合数的质因数分解 + 递归求等比数列前n项和

质因数分解

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;}

等比数列求和（从1开始）

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;}