P5107 能量采集

传送门

官方题解

话说最后的答案忘记取模了结果连暴力都挂了可海星……

//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
    R int res,f=1;R char ch;
    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
    return res*f;
}
char sr[1<<21],z[20];int C=-1,Z=0;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
void print(R int x){
    if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
    while(z[++Z]=x%10+48,x/=10);
    while(sr[++C]=z[Z],--Z);sr[++C]='
';
}
const int N=55,P=998244353,G=5,K=230;
inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}
inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
inline int ksm(R int x,R int y){
    int res=1;
    for(;y;y>>=1,x=1ll*x*x%P)if(y&1)res=1ll*res*x%P;
    return res;
}
struct Matrix{
    int a[N][N],n,m;
    Matrix(){memset(a,0,sizeof(a)),n=m=0;}
    inline int* operator [](R int x){return a[x];}
    inline void operator =(Matrix b){n=b.n,m=b.m;fp(i,1,n)fp(j,1,m)a[i][j]=b[i][j];}
    Matrix operator *(Matrix b){
        Matrix res;res.n=n,res.m=b.m;
        fp(i,1,n)fp(j,1,b.m)fp(k,1,m)res[i][j]=add(res[i][j],1ll*a[i][k]*b[k][j]%P);
    	return res;
    }
}Base,A,Ans,f[G+1][K+1];int n,m,deg[N],q,u,v,a[N],head[N],k,tot,x,sum;
struct eg{int v,nx;}e[N*N];
inline void add_edge(R int u,R int v){e[++tot]={v,head[u]},head[u]=tot;}
int main(){
//	freopen("testdata.in","r",stdin);
    n=read(),m=read(),q=read();fp(i,1,n)Base[1][i]=read();Base.n=1,Base.m=n;
    fp(i,1,m)u=read(),v=read(),++deg[u],add_edge(u,v);fp(i,1,n)++deg[i],add_edge(i,i);
    fp(i,1,n)deg[i]=ksm(deg[i],P-2);fp(u,1,n)go(u)A[u][v]=add(A[u][v],deg[u]);A.n=A.m=n;
    fp(g,1,G){
    	fp(i,1,n)f[g][0][i][i]=1;f[g][0].n=f[g][0].m=n;
    	fp(i,1,K-1)f[g][i]=f[g][i-1]*A;
    	A=A*f[g][K-1];
	}while(q--){
		x=read(),Ans=Base,sum=0;
		for(R int i=1;x;++i)Ans=Ans*f[i][x%K],x/=K;
		fp(i,1,n)sum^=Ans[1][i];print(sum%P);
	}return Ot(),0;
}
原文地址:https://www.cnblogs.com/bztMinamoto/p/10138746.html