CRB and Puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1177    Accepted Submission(s): 468

同样的题：不同的写法，却得不出相同的答案。poj3233 Matrix Power Series

求解

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#include<cstdio>
#include<cstring>
using namespace std;
const int mod=2015;
struct matrix{
	int s[102][102];
	matrix(){
		memset(s,0,sizeof s);
	}
}A,F;int n,m,T;int ans;
matrix operator *(const matrix &a,const matrix &b){
	matrix c;
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			for(int k=0;k<n;k++){
				c.s[i][j]+=a.s[i][k]*b.s[k][j];
			}
			c.s[i][j]%=mod;
		}
	}
	return c;
}
matrix fpow(matrix a,int p){
	matrix res;
	for(int i=0;i<n;i++) res.s[i][i]=1;
	for(;p;p>>=1,a=a*a) if(p&1) res=res*a;
	return res;
}
int main(){
	for(scanf("%d",&T);T--;){
		scanf("%d%d",&n,&m);
		matrix A;
		for(int i=0,k,x;i<n;i++){
			scanf("%d",&k);
			while(k--){
				scanf("%d",&x);x--;
				A.s[i][x]=1;
			}
		}
		for(int i=0;i<n;i++) A.s[i][i+n]=A.s[i+n][i+n]=1;
		n<<=1;
		A=fpow(A,m);
		n>>=1;
		ans=1;
		for(int i=0;i<n;i++){
        	for(int j=n;j<2*n;j++){
        		ans+=A.s[i][j];
			}
		}
		ans%=mod;
		printf("%d\n",ans);
	}
	return 0;
}

hdu5411 CRB and Puzzle

原文：http://www.cnblogs.com/shenben/p/6726629.html