NC20358. [SDOI2013]方程
描述
输入描述
输入含有多组数据，第一行两个正整数T，p。T表示这个测试点内的数据组数，p的含义见题目描述。对于每组数据，第一行四个非负整数n，n1，n2，m。第二行n1+n2个正整数，表示A1..n1+n2。请注意，如果n1+n2等于0，那么这一行会成为一个空行。
输出描述
共T行，每行一个正整数表示取模后的答案。
示例1
输入:
3 10007 3 1 1 6 3 3 3 0 0 5 3 1 1 3 3 3
输出:
3 6 0
说明:
【样例说明】C++14(g++5.4) 解法, 执行用时: 92ms, 内存消耗: 4456K, 提交时间: 2020-02-19 10:39:28
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 100010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
ll c, f(1);
for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
for(;isdigit(c);c=getchar())x=x*10+c-0x30;
return f*x;
}
ll mod;
struct EasyMath
{
ll prime[maxn], phi[maxn], mu[maxn];
bool mark[maxn];
ll fastpow(ll a, ll b, ll c)
{
ll t(a%c), ans(1ll);
for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;
return ans;
}
void exgcd(ll a, ll b, ll &x, ll &y)
{
if(!b){x=1,y=0;return;}
ll xx, yy;
exgcd(b,a%b,xx,yy);
x=yy, y=xx-a/b*yy;
}
ll inv(ll x, ll p) //p是素数
{return fastpow(x%p,p-2,p);}
ll inv2(ll a, ll p)
{
ll x, y;
exgcd(a,p,x,y);
return (x+p)%p;
}
void shai(ll N)
{
ll i, j;
for(i=2;i<=N;i++)mark[i]=false;
*prime=0;
phi[1]=mu[1]=1;
for(i=2;i<=N;i++)
{
if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;
for(j=1;j<=*prime and i*prime[j]<=N;j++)
{
mark[i*prime[j]]=true;
if(i%prime[j]==0)
{
phi[i*prime[j]]=phi[i]*prime[j];
break;
}
mu[i*prime[j]]=-mu[i];
phi[i*prime[j]]=phi[i]*(prime[j]-1);
}
}
}
ll CRT(vector<ll> a, vector<ll> m) //要求模数两两互质
{
ll M=1, ans=0, n=a.size(), i;
for(i=0;i<n;i++)M*=m[i];
for(i=0;i<n;i++)(ans+=a[i]*(M/m[i])%M*inv2(M/m[i],m[i]))%=M;
return ans;
}
}em;
struct CombinatorialNumber_mod
{
ll p[20], q[20], fact[20][maxn], tot, t[20];
void init(ll P)
{
tot=0;
ll i, j;
for(i=2;i*i<=P;i++)
if(P%i==0)
{
p[++tot]=i;
q[tot]=0;
while(P%i==0)q[tot]++, P/=i;
}
if(P>1)p[++tot]=P, q[tot]=1;
rep(i,1,tot)
{
fact[i][0]=1;
t[i]=1;
rep(j,1,q[i])t[i]*=p[i];
rep(j,1,maxn-1)
{
if(j%p[i]==0)fact[i][j]=fact[i][j-1];
else fact[i][j]=fact[i][j-1]*j%t[i];
}
}
}
pll fact_mod(ll n, ll id)
{
if(n<p[id])return pll(0,fact[id][n]);
pll ans = pll( n/p[id], em.fastpow(fact[id][t[id]],n/t[id],t[id]) );
(ans.second*=fact[id][n%t[id]])%t[id];
auto nex = fact_mod(n/p[id],id);
ans.first+=nex.first;
(ans.second*=nex.second)%=t[id];
return ans;
}
ll exlucas(ll n, ll m, ll id)
{
ll cnt;
auto a=fact_mod(n,id), b=fact_mod(m,id), c=fact_mod(n-m,id);
a.first-=b.first+c.first;
(a.second*=em.inv2(b.second*c.second%t[id],t[id]))%t[id];
return em.fastpow(p[id],a.first,t[id])*a.second%t[id];
}
ll calc(ll n, ll m)
{
if(m>n or m<0 or n<0)return 0;
vector<ll> a, v(t+1,t+tot+1);
ll i;
rep(i,1,tot)a.emb(exlucas(n,m,i));
return em.CRT(a,v);
}
}Cmod;
ll ans, n1, n2, n, M, a[20];
void dfs(ll pos, ll f, ll sm)
{
if(pos>n1)
{
( ans += f*Cmod.calc(M-sm-1,n-1) )%=mod;
return;
}
dfs(pos+1,f,sm);
dfs(pos+1,-f,sm+a[pos]);
}
int main()
{
ll T, i, s;
T=read(), mod=read();
Cmod.init(mod);
while(T--)
{
n=read(), n1=read(), n2=read(), M=read();
rep(i,1,n1)a[i]=read();
s=0;
rep(i,n1+1,n1+n2)a[i]=read(), s+=a[i]-1;
ans=0;
dfs(1,1,s);
printf("%lld\n",(ans+mod)%mod);
}
return 0;
}
C++11(clang++ 3.9) 解法, 执行用时: 289ms, 内存消耗: 584K, 提交时间: 2020-09-19 12:21:24
#include<bits/stdc++.h>
#define Min(x,y) ((x)<(y)?(x):(y))
#define Max(x,y) ((x)>(y)?(x):(y))
using namespace std;
typedef long long ll;
int T,n1,n2;
ll n,m,p,a[20],f[1000005];
template <class T>
void read(T &x)
{
char c; int sign=1;
while((c=getchar())>'9'||c<'0') if(c=='-') sign=-1; x=c-48;
while((c=getchar())>='0'&&c<='9') x=(x<<1)+(x<<3)+c-48; x*=sign;
}
ll quickpow(ll a,ll b,ll mod)
{
ll ret=1;
while(b)
{
if(b&1) ret=ret*a%mod;
a=a*a%mod;
b>>=1;
}
return ret;
}
ll exgcd(ll a,ll b,ll &x,ll &y)
{
if(!b) { x=1; y=0; return a; }
ll d=exgcd(b,a%b,x,y);
ll t=x; x=y; y=t-a/b*y;
return d;
}
ll inv(ll a,ll p)
{
ll x,y;
exgcd(a,p,x,y);
return (x%p+p)%p;
}
ll fac(ll n,ll pi,ll pk)
{
if(!n) return 1LL;
if(n<pi) return f[n];
return quickpow(f[pk-1],n/pk,pk)*f[n%pk]%pk*fac(n/pi,pi,pk)%pk;
}
ll C(ll n,ll m,ll pi,ll pk)
{
if(n<m) return 0;
f[0]=1;
for(int i=1;i<=pk;i++)
if(i%pi!=0) f[i]=f[i-1]*i%pk;
else f[i]=f[i-1];
ll jn=fac(n,pi,pk),jm=fac(m,pi,pk),jnm=fac(n-m,pi,pk);
int k=0;
for(ll i=n;i;i/=pi) k+=i/pi;
for(ll i=m;i;i/=pi) k-=i/pi;
for(ll i=n-m;i;i/=pi) k-=i/pi;
return jn * inv(jm,pk)%pk * inv(jnm,pk)%pk * quickpow(pi,k,pk)%pk;
}
ll crt(ll a,ll pk)
{
ll x=p/pk;
return a*x%p*inv(x,pk)%p;//关于pk的逆元
}
ll solve(ll n,ll m,ll pi,ll pk)
{
ll ret=0;
for(int i=0,t=(1<<n1);i<t;++i)
{
int opt=1;
ll nown=n;
for(int j=0;j<n1;++j)
if(i>>j&1)
opt=-opt,nown-=a[j+1];
ret=(ret+opt*C(nown,m,pi,pk))%pk;
}
return ret;
}
ll exlucas(ll n,ll m,ll P)
{
if(n<m) return 0;
ll ret=0;
for(ll i=2;i*i<=P;++i)
{
if(P%i==0)
{
ll pk=1;
while(P%i==0)
{
pk*=i;
P/=i;
}
ret=(ret+crt(solve(n,m,i,pk),pk))%p;
}
}
if(P!=1) ret=(ret+crt(solve(n,m,P,P),P))%p;
return (ret%p+p)%p;
}
int main()
{
read(T);read(p);
while(T--)
{
read(n);read(n1);read(n2);read(m);
for(int i=1;i<=n1+n2;++i) read(a[i]);
for(int i=n1+1;i<=n1+n2;++i) m-=(a[i]-1);
printf("%lld\n",exlucas(m-1,n-1,p));
}
return 0;
}