记录编号 |
584898 |
评测结果 |
AAAAAAAAAA |
题目名称 |
一二三四五 |
最终得分 |
100 |
用户昵称 |
┭┮﹏┭┮ |
是否通过 |
通过 |
代码语言 |
C++ |
运行时间 |
0.000 s |
提交时间 |
2023-11-16 20:22:11 |
内存使用 |
0.00 MiB |
显示代码纯文本
#include <bits/stdc++.h>
using namespace std;
//递推式子 + 矩阵快速幂
typedef long long ll;
const int mod = 1e5+7;
int n;
struct Matrix{
ll a[6][6],n,m;
void clear(){n = m = 0;memset(a,0,sizeof(a));}
Matrix operator * (const Matrix &x)const{
Matrix y;y.clear();
y.n = n,y.m = m;
for(int i = 0;i < n;i++)
for(int j = 0;j < m;j++)
for(int k = 0;k < m;k++)
y.a[i][j] = (y.a[i][j] + (a[i][k] * x.a[k][j]) % mod) % mod;
return y;
}
}c,f;
int main(){
freopen("five.in","r",stdin);
freopen("five.out","w",stdout);
scanf("%d",&n);
if(n == 1){
printf("2\n");
return 0;
}
n -= 1;
f.n = 1,f.m = 6;
f.a[0][0] = 1,f.a[0][1] = 2,f.a[0][2] = 2;
c.n = c.m = 6;
c.a[0][0] = 1,c.a[1][0] = 2,c.a[2][0] = 2;
c.a[0][1] = 2,c.a[1][1] = 1,c.a[2][1] = 2;
c.a[0][2] = 2,c.a[1][2] = 2,c.a[2][2] = 1;
c.a[0][3] = 1,c.a[1][4] = 1,c.a[2][5] = 1;
while(n){
if(n & 1)f = f * c;
c = c * c;
n >>= 1;
}
printf("%lld\n",f.a[0][1]);
return 0;
}