P2480 [SDOI2010]古代猪文

题意：

给你n和g，求$g^{{\sum }_{d|n}{C}_n^d}\mathrm{m}\mathrm{o}\mathrm{d}p$
p=999911659

题解：

这个一个综合性很强的数论题
涉及到欧拉定理，Lucas定理，中国剩余定理，挺好的一个题
首先根据欧拉定理推论：

若正整数a，n互质，对于任意的正整数b，有$a^b\equiv a^{b\mathrm{m}\mathrm{o}\mathrm{d}\varphi (n)}(\mathrm{m}\mathrm{o}\mathrm{d}n)$

所以
$=g^{{\sum }_{d|n}{C}_n^d\mathrm{m}\mathrm{o}\mathrm{d}\varphi (p)}\mathrm{m}\mathrm{o}\mathrm{d}p$
$\varphi (p)=p-1=999911658$
$=g^{{\sum }_{d|n}{C}_n^d\mathrm{m}\mathrm{o}\mathrm{d}999911658}\mathrm{m}\mathrm{o}\mathrm{d}p$
现在的关键在于求${\sum }_{d|n}{C}_n^d\mathrm{m}\mathrm{o}\mathrm{d}999911658$,
999911658不是质数，咋搞？那我们可以将其质因子分解：999911658=4679 * 3 * 2 *35617，每个质因子的次数都是1，所以我们只需要用CRT来求解如下的方程组

求出p后，最后再一个快速幂输出答案

CRT：

inline ll CRT()
{ll ans=0;for(register int i=1;i<=cnt;i++){ll M=mod/p[i],t=qpow(M,p[i]-2,p[i]);ans=(ans+a[i]%mod*t%mod*M%mod)%mod;}return (ans+mod)%mod;
}

Lucas组合数：

ll C(int a, int b, ll mod)
{if (b > a)return 0;return fac[a] % mod * inv[b] % mod * inv[a - b] % mod;
}
ll Lucas(ll n, ll m, ll p)
{if (m == 0)return 1;return Lucas(n / p, m / p, p) * C(n % p, m % p, p) % p;
}

递推求阶乘逆元

$(n-1)!\times n[n!{]}^{-1}\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}p$

void init() {fact[0] = 1;for (int i = 1; i < maxn; i++) {fact[i] = fact[i - 1] * i %mod;}inv[maxn - 1] = power(fact[maxn - 1], mod - 2);for (int i = maxn - 2; i >= 0; i--) {inv[i] = inv[i + 1] * (i + 1) %mod;}
}

代码：

but，代码存在问题，还没修改出哪里错了

目前95分，错了第一个点，人傻了

// Problem: P2480 [SDOI2010]古代猪文
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P2480
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// Data:2021-08-26 15:50:36
// By Jozky#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{x= 0;char c= getchar();bool flag= 0;while (c < '0' || c > '9')flag|= (c == '-'), c= getchar();while (c >= '0' && c <= '9')x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();if (flag)x= -x;read(Ar...);
}
template <typename T> inline void write(T x)
{if (x < 0) {x= ~(x - 1);putchar('-');}if (x > 9)write(x / 10);putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef LOCALstartTime= clock();freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCALendTime= clock();printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
const int maxn= 4e4 + 9;
ll n, g;
ll fac[maxn];
ll inv[maxn];
const int mod= 999911659;
ll p[maxn];
ll c[maxn];
ll a[maxn];
int cnt= 0;
int tot= 0;
ll poww(ll a, ll b, ll mod)
{ll ans= 1;while (b) {if (b & 1)ans= ans * a % mod;a= a * a % mod;b>>= 1;}return ans % mod;
}
ll exgcd(int a, int b, ll &x, ll &y)
{if (b == 0) {x= 1;y= 0;return a;}int gcd= exgcd(b, a % b, x, y);ll t= x;x= y;y= t - a / b * y;return gcd;
}
ll CRT(int k, ll a[], ll r[], ll mod)
{ll n= 1, ans= 0;ll x,y;for (int i= 1; i <= k; i++)n= n * r[i];for (int i= 1; i <= k; i++) {ll m= n / r[i];exgcd(m, r[i], x, y);ans= (ans + a[i] % mod * m % mod * x % mod) % mod;}return (ans % mod + mod) % mod;
}
ll C(int a, int b, ll mod)
{if (b > a)return 0;return fac[a] % mod * inv[b] % mod * inv[a - b] % mod;
}
ll Lucas(ll n, ll m, ll p)
{if (m == 0)return 1;return Lucas(n / p, m / p, p) * C(n % p, m % p, p) % p;
}
void init(int p)
{fac[0]= 1;for (int i= 1; i < p; i++) {fac[i]= fac[i - 1] * i % p;}inv[p]=0;inv[p-1]=poww(fac[p-1],p-2,p);for(register int i=p-2;i>=0;i--)inv[i]=inv[i+1]*(i+1)%p;
}
void calc(int x)
{init(p[x]);for (int i= 1; i <= tot; i++) {a[x]= (a[x] + Lucas(n, c[i], p[x])) % p[x];}
}int main()
{//rd_test();cin >> n >> g;if (g % mod == 0) {printf("0--\n");return 0;}ll phi= mod - 1;for (int i= 2; i * i <= (mod - 1); i++) { //对mod-1进行质因子分解if (phi % i == 0) {p[++cnt]= i;while (phi % i == 0)phi= phi / i;}}if (phi != 1)p[++cnt]= phi;for (int i= 1; i * i <= n; i++) { //c存的是n的因子if (n % i == 0) {c[++tot]= i;if (i * i != n)c[++tot]= n / i;}}for (int i= 1; i <= cnt; i++)calc(i); //预初理出组合数情况ll sum= CRT(cnt, a, p, mod - 1)%mod;printf("%lld\n", poww(g, sum, mod) % mod);//Time_test();
}

AC代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
#define Mod 999911659
#define mod 999911658
#define maxn 40005
typedef long long ll;
ll n,g;
ll d[maxn],tot;
ll p[10],cnt;inline ll qpow(ll a,ll k,ll p)
{ll res=1;while(k){if(k&1) res=(res*a)%p;a=(a*a)%p;k>>=1;}return res%p;
}ll fac[maxn],inv[maxn];
inline void init(ll p)
{fac[0]=1;for(register int i=1;i<p;i++)fac[i]=fac[i-1]*i%p;inv[p]=0;inv[p-1]=qpow(fac[p-1],p-2,p);for(register int i=p-2;i>=0;i--)inv[i]=inv[i+1]*(i+1)%p;
}inline ll C(ll n,ll m,ll p)
{if(m>n) return 0;return fac[n]*inv[m]%p*inv[n-m]%p;
}inline ll Lucas(ll n,ll m,ll p)
{if(m==0) return 1;return Lucas(n/p,m/p,p)*C(n%p,m%p,p)%p;
}ll a[10];
inline void calc(int x)
{init(p[x]);for(register int i=1;i<=tot;i++)a[x]=(a[x]+Lucas(n,d[i],p[x]))%p[x];
}inline ll CRT()
{ll ans=0;for(register int i=1;i<=cnt;i++){ll M=mod/p[i],t=qpow(M,p[i]-2,p[i]);ans=(ans+a[i]%mod*t%mod*M%mod)%mod;}return (ans+mod)%mod;
}int main()
{scanf("%lld%lld",&n,&g);if(g%Mod==0){printf("0\n");return 0;}ll t=mod;for(register int i=2;i*i<=mod;i++){if(t%i==0){p[++cnt]=i;while(t%i==0) t=t/i;}}if(t!=1) p[++cnt]=t;for(register int i=1;i*i<=n;i++){if(n%i==0){d[++tot]=i;if(i*i!=n) d[++tot]=n/i;}}for(register int i=1;i<=cnt;i++) calc(i);printf("%lld",qpow(g,CRT(),Mod));return 0;
}