单位根反演题单

单位根反演。

单位根反演+矩阵乘法。

求图上路径长度为k的倍数的方案数。
单位根反演+矩阵乘法。

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
ll n;
int m,k,mods;
int upd(int x,int y){ return x+y>=mods?x+y-mods:x+y; }
int quick_pow(int x,int y)
{int ret=1;for (;y;y>>=1){if (y&1) ret=1ll*ret*x%mods;x=1ll*x*x%mods;}return ret;
}
int flag[MAXN],prime[MAXN],a[MAXN],pnum=0,cnt=0;
void Prime_Init(int n)
{flag[1]=1;for (int i=2;i<=n;i++){if (!flag[i]) prime[++cnt]=i;for (int j=1;prime[j]*i<=n&&j<=cnt;j++) {flag[prime[j]*i]=1;if (!(i%prime[j])) break;}}
}
void findp(int x)
{x--,pnum=0;for (int i=1;prime[i]*prime[i]<=x;i++)if (!(x%prime[i])){a[++pnum]=prime[i];while (!(x%prime[i])) x/=prime[i];}
}
int getwn(int x)
{findp(x);for (int i=2;;i++){bool flag=1;for (int j=1;j<=pnum;j++){int t=(x-1)/a[j];if (quick_pow(i,t)==1) { flag=0; break; } }if (flag) return i;}
}struct Matrix
{int n,A[5][5];void init(){ for (int i=0;i<n;i++) A[i][i]=1; }Matrix(int n1=0){n=n1;memset(A,0,sizeof A);}Matrix operator * (const Matrix &b){Matrix Ans(n);for (int i=0;i<n;i++)for (int j=0;j<n;j++)for (int k=0;k<n;k++) Ans.A[i][j]=upd(Ans.A[i][j],1ll*A[i][k]*b.A[k][j]%mods);return Ans;}Matrix operator + (const Matrix &b){Matrix Ans(n);for (int i=0;i<n;i++)for (int j=0;j<n;j++) Ans.A[i][j]=upd(Ans.A[i][j],(A[i][j]+b.A[i][j])%mods);return Ans;}Matrix operator ^ (const int &b){Matrix ret(n),x=*this;ret.init();for (ll y=b;y;y>>=1){if (y&1) ret=ret*x;x=x*x;}return ret;}void print(){for (int i=0;i<n;i++){for (int j=0;j<n;j++) cout<<setw(4)<<A[i][j];cout<<endl;}cout<<endl;}
};
void build(Matrix &f,Matrix &G,int w)
{for (int i=0;i<m;i++)for (int j=0;j<m;j++) f.A[i][j]=1ll*G.A[i][j]*w%mods;for (int i=0;i<m;i++) f.A[i][i]=upd(f.A[i][i],1);
}
int main()
{Prime_Init(100000);while (scanf("%d%lld",&m,&n)!=EOF){k=read(),mods=read();int l=read(),s=read()-1,t=read()-1;Matrix f(m),Ans(m),G(m);for (int i=1;i<=l;i++){int u=read()-1,v=read()-1;G.A[u][v]++;}int wn=getwn(mods),ans=0;wn=quick_pow(wn,(mods-1)/k);
//		cout<<wn<<endl;for (int i=0,w=1;i<k;i++,w=1ll*w*wn%mods){build(f,G,w);
//			f.print();Ans=f^n;
//			Ans.print();ans=upd(ans,Ans.A[s][t]);}printf("%d\n",1ll*ans*quick_pow(k,mods-2)%mods);}return 0;
}