[题目链接]
思路:
构建一个6*6的转移矩阵,然后矩阵快速幂进行求解。
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int Max_n=110;
ll n=6; //n*n矩阵
struct Mat{
ll m[Max_n][Max_n];
Mat(){memset(m,0,sizeof(m));}
Mat operator*(Mat &a){
Mat b;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++)
for(int k=0;k<n;k++)
b.m[i][j]=(b.m[i][j]+m[i][k]*a.m[k][j])%mod;
}
return b;
}
};
Mat quickMod(Mat base,ll exp){
Mat ans;
for(int i=0;i<n;i++)ans.m[i][i]=1; //单位矩阵初始化
while(exp){
if(exp&1)ans=ans*base;
exp>>=1;
base=base*base;
}
return ans;
}
int main()
{
Mat base;
ll mat[6][6]={
{1,1,1,1,1,1},
{1,0,0,0,0,0},
{0,0,1,3,3,1},
{0,0,0,1,2,1},
{0,0,0,0,1,1},
{0,0,0,0,0,1},
};
for(int i=0;i<6;i++){ //初始化base矩阵
for(int j=0;j<6;j++)
base.m[i][j]=mat[i][j];
}
ll t,exp;
scanf("%lld",&t);
while(t--){
scanf("%lld",&exp);
Mat c=quickMod(base,exp-1);
printf("%lld\n",(c.m[0][0]+c.m[0][2]*8+c.m[0][3]*4+c.m[0][4]*2+c.m[0][5])%mod);
}
return 0;
}