第六屆藍橋杯——壘骰子（矩陣快速冪）
阿新 • • 發佈：2019-01-21
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <stack>
using namespace std;
typedef long long ll;
#define PI 3.1415926535897932
#define E 2.718281828459045
#define INF 0x3f3f3f3f
#define mod 1000000007
const int M=1005;
int n,m;
int cnt;
int sx,sy,sz;
int mp[M][M];
int pa[M*10],rankk[M];
int head[M*6],vis[M*10];
int dis[M][10];
ll prime[M*1000];
bool isprime[M*1000];
int lowcost[M],closet[M];
char st1[5050],st2[5050];
int len[M*6];
typedef pair<int ,int> ac;
vector<int> g[M*10];
int dp[M];
int sums[M*10];
int has[105000];
int month[13]= {0,31,59,90,120,151,181,212,243,273,304,334,0};
int dir[8][2]= {{0,1},{0,-1},{-1,0},{1,0},{1,1},{1,-1},{-1,1},{-1,-1}};
void getpri()
{
ll i;
int j;
cnt=0;
memset(isprime,false,sizeof(isprime));
for(i=2; i<1000000LL; i++)
{
if(!isprime[i])prime[cnt++]=i;
for(j=0; j<cnt&&prime[j]*i<1000000LL; j++)
{
isprime[i*prime[j]]=1;
if(i%prime[j]==0)break;
}
}
}
ll qk_mul(ll a,ll b)
{
ll t=0;
while(b)
{
if(b&1)
t=(t+a)%mod;
a=(a<<1)%mod;
b>>=1;
}
t%=mod;
return t;
}
ll qk_mod(ll a,ll b)
{
ll ans=1;
while(b)
{
if(b&1) ans=qk_mul(ans,a);
a=qk_mul(a,a);
b>>=1;
}
ans%=mod;
return ans;
}
int gcd(int a,int b)
{
return b == 0 ? a : gcd(b, a%b);
}
int lcm(int a,int b)
{
return a*b/gcd(a,b);
}
int findx(int t)
{
if(t!=pa[t])
pa[t]=findx(pa[t]);
return pa[t];
}
void unionx(int x,int y)
{
x=findx(x);
y=findx(y);
if(x!=y)
{
pa[y]=x;
}
}
void init()
{
for(int i=0; i<101; i++)
pa[i]=i;
}
int heap[100005];
void push(int x)
{
int i=++sz;
while(i>1) //i>0
{
int p=i/2; //(i-1)/2
if(heap[p]<=x)break;
heap[i]=heap[p];
i=p;
}
heap[i]=x;
/* a[++sz] = x;
int t = a[sz];
int tson = sz;
while( (tson > 1)&&( a[tson/2] > t))
{
a[tson] = a[tson/2];
tson = tson/2;
}
a[tson] = x;
*/
}
int pop()
{
int ret=heap[1];//int ret=a[0]
int x=heap[sz--];//a[--sz]
int i=1;// i=0
while(2*i<sz) //2*i+1
{
int a=i*2,b=i*2+1;
if(b<sz&&heap[b]<heap[a])a=b;
if(heap[a]>=x)break;
heap[i]=heap[a];
i=a;
}
heap[i]=x;
return ret;
}
int lowbit(int i)
{
return i&(-i);
}
void add(int i,int v)//修改值並向父節點修改
{
while(i<=M)
{
sums[i]+=v; //c[i]開始都是0，每經過一個數，它的c[i]+1表示c[i]
i+=lowbit(i);
}
}
int sum(int i)//求和
{
ll s=0;
while(i>=1)
{
s+=sums[i];
i-=lowbit(i);
}
return s;
}
char pre[7][7]=
{
{'>','>','<','<','<','>','>'},
{'>','>','<','<','<','>','>'},
{'>','>','>','>','<','>','>'},
{'>','>','>','>','<','>','>'},
{'<','<','<','<','<','=','0'},
{'>','>','>','>','0','>','>'},
{'<','<','<','<','<','0','='}
};
struct Mat
{
ll mat[6][6];
Mat(int x)
{
memset(mat,0,sizeof(mat));
for(int i=0; i<6; i++)mat[i][i]=x;
}
};
Mat operator*(const Mat& a,const Mat& b)
{
Mat c(0);
int i,j,k;
for(i=0; i<6; i++)
for(j=0; j<6; j++)
for(k=0; k<6; k++)
{
c.mat[i][j]=(c.mat[i][j]+a.mat[i][k]*b.mat[k][j])%mod;
}
return c;
}
Mat qmod(Mat x,int b)
{
Mat res(1);
while(b)
{
if(b&1)res=res*x;
x=x*x;
b>>=1;
}
return res;
}
ll qmod2(ll x,int b)
{
ll res=1;
while(b)
{
if(b&1)res=res*x%mod;
x=x*x%mod;
b>>=1;
}
return res;
}
int main()
{
int i,j,t,u,v;
Mat ans(1);
scanf("%d%d",&n,&m);
for(i=0;i<6;i++)
for(j=0;j<6;j++)
ans.mat[i][j]=1;
for(i=0;i<m;i++)
{
scanf("%d%d",&u,&v);
ans.mat[u-1][(v+2)%6]=0;
ans.mat[v-1][(u+2)%6]=0;
}
Mat res=qmod(ans,n-1);//注意冪的值！！
ll op=0;
for(i=0;i<6;i++)
for(j=0;j<6;j++)
op=(op+res.mat[i][j])%mod;
op=(op*qmod2(4,n))%mod;
printf("%I64d\n",op);
return 0;
}
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <stack>
using namespace std;
typedef long long ll;
#define PI 3.1415926535897932
#define E 2.718281828459045
#define INF 0x3f3f3f3f
#define mod 1000000007
const int M=1005;
int n,m;
int cnt;
int sx,sy,sz;
int mp[M][M];
int pa[M*10],rankk[M];
int head[M*6],vis[M*10];
int dis[M][10];
ll prime[M*1000];
bool isprime[M*1000];
int lowcost[M],closet[M];
char st1[5050],st2[5050];
int len[M*6];
typedef pair<int ,int> ac;
vector<int> g[M*10];
int dp[M];
int sums[M*10];
int has[105000];
int month[13]= {0,31,59,90,120,151,181,212,243,273,304,334,0};
int dir[8][2]= {{0,1},{0,-1},{-1,0},{1,0},{1,1},{1,-1},{-1,1},{-1,-1}};
void getpri()
{
ll i;
int j;
cnt=0;
memset(isprime,false,sizeof(isprime));
for(i=2; i<1000000LL; i++)
{
if(!isprime[i])prime[cnt++]=i;
for(j=0; j<cnt&&prime[j]*i<1000000LL; j++)
{
isprime[i*prime[j]]=1;
if(i%prime[j]==0)break;
}
}
}
ll qk_mul(ll a,ll b)
{
ll t=0;
while(b)
{
if(b&1)
t=(t+a)%mod;
a=(a<<1)%mod;
b>>=1;
}
t%=mod;
return t;
}
ll qk_mod(ll a,ll b)
{
ll ans=1;
while(b)
{
if(b&1) ans=qk_mul(ans,a);
a=qk_mul(a,a);
b>>=1;
}
ans%=mod;
return ans;
}
int gcd(int a,int b)
{
return b == 0 ? a : gcd(b, a%b);
}
int lcm(int a,int b)
{
return a*b/gcd(a,b);
}
int findx(int t)
{
if(t!=pa[t])
pa[t]=findx(pa[t]);
return pa[t];
}
void unionx(int x,int y)
{
x=findx(x);
y=findx(y);
if(x!=y)
{
pa[y]=x;
}
}
void init()
{
for(int i=0; i<101; i++)
pa[i]=i;
}
int heap[100005];
void push(int x)
{
int i=++sz;
while(i>1) //i>0
{
int p=i/2; //(i-1)/2
if(heap[p]<=x)break;
heap[i]=heap[p];
i=p;
}
heap[i]=x;
/* a[++sz] = x;
int t = a[sz];
int tson = sz;
while( (tson > 1)&&( a[tson/2] > t))
{
a[tson] = a[tson/2];
tson = tson/2;
}
a[tson] = x;
*/
}
int pop()
{
int ret=heap[1];//int ret=a[0]
int x=heap[sz--];//a[--sz]
int i=1;// i=0
while(2*i<sz) //2*i+1
{
int a=i*2,b=i*2+1;
if(b<sz&&heap[b]<heap[a])a=b;
if(heap[a]>=x)break;
heap[i]=heap[a];
i=a;
}
heap[i]=x;
return ret;
}
int lowbit(int i)
{
return i&(-i);
}
void add(int i,int v)//修改值並向父節點修改
{
while(i<=M)
{
sums[i]+=v; //c[i]開始都是0，每經過一個數，它的c[i]+1表示c[i]
i+=lowbit(i);
}
}
int sum(int i)//求和
{
ll s=0;
while(i>=1)
{
s+=sums[i];
i-=lowbit(i);
}
return s;
}
char pre[7][7]=
{
{'>','>','<','<','<','>','>'},
{'>','>','<','<','<','>','>'},
{'>','>','>','>','<','>','>'},
{'>','>','>','>','<','>','>'},
{'<','<','<','<','<','=','0'},
{'>','>','>','>','0','>','>'},
{'<','<','<','<','<','0','='}
};
struct Mat
{
ll mat[6][6];
Mat(int x)
{
memset(mat,0,sizeof(mat));
for(int i=0; i<6; i++)mat[i][i]=x;
}
};
Mat operator*(const Mat& a,const Mat& b)
{
Mat c(0);
int i,j,k;
for(i=0; i<6; i++)
for(j=0; j<6; j++)
for(k=0; k<6; k++)
{
c.mat[i][j]=(c.mat[i][j]+a.mat[i][k]*b.mat[k][j])%mod;
}
return c;
}
Mat qmod(Mat x,int b)
{
Mat res(1);
while(b)
{
if(b&1)res=res*x;
x=x*x;
b>>=1;
}
return res;
}
ll qmod2(ll x,int b)
{
ll res=1;
while(b)
{
if(b&1)res=res*x%mod;
x=x*x%mod;
b>>=1;
}
return res;
}
int main()
{
int i,j,t,u,v;
Mat ans(1);
scanf("%d%d",&n,&m);
for(i=0;i<6;i++)
for(j=0;j<6;j++)
ans.mat[i][j]=1;
for(i=0;i<m;i++)
{
scanf("%d%d",&u,&v);
ans.mat[u-1][(v+2)%6]=0;
ans.mat[v-1][(u+2)%6]=0;
}
Mat res=qmod(ans,n-1);//注意冪的值！！
ll op=0;
for(i=0;i<6;i++)
for(j=0;j<6;j++)
op=(op+res.mat[i][j])%mod;
op=(op*qmod2(4,n))%mod;
printf("%I64d\n",op);
return 0;
}