poj1845（有错）

数论知识知，一个数的因子和函数sigma（n）有如下定义：

其中

而又知

所以

且

而又知

注意前提！！b，c要互质。

如果不互质呢？没想出怎么办。我估计我就错在这里了。

网上有这个公式

额扯淡了，高精度直接给跪。。难不成还得写个高精度求模么- -。。

然后

这么推是没错的，但是乘法逆就没办法了，这也就是discuss里面有几组数据为什么一直不过的原因。

59407 1
2
217823 1
2
396041 1
2
59407 2
3
59407 59407
2

错误代码：

#include <iostream>
#include <math.h>

using namespace std;

#define LL long long int
#define mod 9901
#define MAXN 5000000
#define setbitzero(a) (isprime[(a)>>5]&=(~(1<<((a)&31))))
#define setbitone(a) (isprime[(a)>>5]|=(1<<((a)&31)))
#define ISPRIME(a) (isprime[(a)>>5]&(1<<((a)&31)))

LL plist[6000000], pcount;
LL isprime[(MAXN >> 5) + 1];
LL a, b, n, f[10000], nf[10000];

void initprime()
{
    LL i, j, m;
    LL t = (MAXN >> 5) + 1;
    for (i = 0; i < t; ++i)isprime[i] = 2863311530;
    plist[0] = 2;
    setbitone(2);
    setbitzero(1);
    m = (LL) sqrt(MAXN);
    for (pcount = 1, i = 3; i <= m; i += 2)
        if (ISPRIME(i))
            for (plist[pcount++] = i, j = i << 1; j <= MAXN; j += i)
                setbitzero(j);
    if (!(i & 1))++i;
    for (; i <= MAXN; i += 2)if (ISPRIME(i))plist[pcount++] = i;
}
int prime_factor(LL n, LL* f, LL *nf)
{
    int cnt = 0;
    int n2 = sqrt((double) n);
    for (int i = 0; n > 1 && plist[i] <= n2; ++i)
        if (n % plist[i] == 0)
        {
            for (nf[cnt] = 0; n % plist[i] == 0; ++nf[cnt], n /= plist[i]);
            f[cnt++] = plist[i];
        }
    if (n > 1) nf[cnt] = 1, f[cnt++] = n;
    return cnt;
}
//扩展欧几里得
void gcd(LL a, LL b, LL &d, LL &x, LL &y)
{
    if (!b)
    {
        d = a;
        x = 1;
        y = 0;
    }
    else
    {
        gcd(b, a%b, d, y, x);
        y -= x*(a / b);
    }
}
//a关于n的乘法逆
LL inv(LL a, LL n)
{
    LL d, x, y;
    gcd(a, n, d, x, y);
    return d == 1 ? (x + n) % n : -1;
}
//计算a^b % n
LL modexp_recursion(LL a, LL b, LL n)
{
    LL t = 1;
    if (b == 0)
        return 1;
    if (b == 1)
        return a%n;
    t = modexp_recursion(a, b >> 1, n);
    t = t*t % n;
    if (b & 0x1)
        t = t*a % n;
    return t;
}

int main()
{
    initprime();
    while (cin >> a >> b)
    {
        int num = prime_factor(a, f, nf);
        
        LL ans = 1;
        
        for (int i = 0; i < num; i++)
        {
            ans *= ((modexp_recursion(f[i], nf[i] * b + 1,mod) - 1) * (inv(f[i] - 1, mod) % mod) + mod) % mod;
        }
        
        cout << ans % mod << endl;
    }
}

A^B = p1^(a1*B) * p2^(a2*B) *...* pn^(an*B);

ans = [1+p1+p1^2+...+p1^(a1*B)] * [1+p2+p2^2+...+p2^(a2*B)] *...* [1+pn+pn^2+...+pn^(an*B)].

等比求和公式 (p^(n+1)-1)/(p-1)，由于这里有1/(p-1)，需要求(p-1)关于mod的逆元 但是，存在逆元需要 gcd(a,m)  = 1。所以直接二分+快速求幂。

然后代码也是异常简洁：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>

using namespace std;

#define lint __int64
#define MAXN 100000
#define M 9901
struct Factor { lint base, exp; };
Factor f[MAXN]; lint fn;
lint p[MAXN], a[MAXN], pn;

void prime ()
{
    int i, j; pn = 0;
    memset(a,0,sizeof(a));
    for ( i = 2; i < MAXN; i++ )
    {
        if ( !a[i] ) p[pn++] = i;
        for ( j = 0; j < pn && i * p[j] < MAXN && (p[j] <= a[i] || a[i] == 0); j++ )
            a[i*p[j]] = p[j];
    }
}

void Factorization ( int num )
{
    fn = 0;
    for ( int i = 0; i < pn && p[i] <= num; i++ )
    {
        if ( num % p[i] ) continue;
        f[++fn].base = p[i];
        f[fn].exp = 0;
        while ( num % p[i] == 0 )
        {
            f[fn].exp++;
            num /= p[i];
        }
    }
    if ( num != 1 )
    {
        f[++fn].base = num;
        f[fn].exp = 1;
    }
}

int mod_exp ( int a, lint b )
{
    int ret = 1;
    a = a % M;
    while ( b >= 1 )
    {
        if ( b & 1 ) ret = ret * a % M;
        a = a * a % M;
        b >>= 1;
    }
    return ret;
}

int sum_exp ( int p, lint exp )
{
    if ( exp == 0 ) return 1;
    lint tmp1, tmp2, mid = exp / 2;
    if ( exp & 1 )
    {
        tmp1 = sum_exp (p, mid);
        tmp2 = mod_exp (p, mid + 1);
        return (tmp1+tmp2*tmp1) % M;
    }
    else
    {
        tmp1 = sum_exp (p, mid-1);
        tmp2 = mod_exp (p, mid);
        return (tmp1 + tmp2 + p*tmp2*tmp1) % M;
    }
}

int main()
{
    prime();
    int A, B, ret = 1;
    scanf("%d%d",&A,&B);
    if ( A == 0 ) {printf("0\n");return 0;}
    if ( B == 0 || A == 1 ) {printf("1\n"); return 0;}
    Factorization ( A );
    for ( int i = 1; i <= fn; i++ )
        ret = ret * sum_exp(f[i].base, f[i].exp*B) % M;
    printf("%d\n",ret);
}

好不容易熬到周日了，该休息休息了- -。。