D - A Short problem HDU - 4291——矩阵快速幂

Think:

1知识点：矩阵快速幂

2反思：

1>内嵌关系取模考虑循环节和矩阵快速幂

2>矩阵快速幂通过把数放到初始矩阵的不同位置，进而将普通的递推式转换为“矩阵的等比数列”，进而快速幂求解递推式

3>矩阵快速幂解题步骤:

第一步列出递推式

第二步建立矩阵递推式，找到转移矩阵

vjudge题目链接

以下为Accepted代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int mod0 = 1000000007;
const int mod1 = 222222224;
const int mod2 = 183120;
const int mod3 = 240;

struct Mat{
LL rx[4][4];
};

struct Mat Mul(struct Mat a, struct Mat b, int mod);
struct Mat pow_mod(struct Mat a, int k, int mod);

/*循环节*/
/*int main(){
LL a, b, c;
a = 0, b = 1;
for(LL i = 3; ; i++){
c = (3*b + a)%mod;///通过mod得到mod1进而得到mod2进而得到mod3
a = b;
b = c;
if(a == 0 && b == 1){
printf("%lld\n", i-2);
break;
}
}
return 0;
}*/
/*mod0 = 1000000007*/
/*mod1 = 222222224*/
/*mod2 = 183120*/
/*mod3 = 240*/

int main(){
struct Mat tmp;
tmp.rx[0][0] = 3;
tmp.rx[0][1] = 1;
tmp.rx[1][0] = 1;
tmp.rx[1][1] = 0;
LL n;
while(~scanf("%lld", &n)){
n %= mod3;

if(!n){
printf("0\n");
continue;
}

n = pow_mod(tmp, n-1, mod2).rx[0][0];

n = pow_mod(tmp, n-1, mod1).rx[0][0];

n = pow_mod(tmp, n-1, mod0).rx[0][0];

printf("%lld\n", n);
}
return 0;
}
struct Mat pow_mod(struct Mat a, int k, int mod){
struct Mat ans;
for(int i = 0; i < 2; i++){
ans.rx[i][i] = 1;
for(int j = i+1; j < 2; j++){
ans.rx[i][j] = ans.rx[j][i] = 0;
}
}
while(k){
if(k & 1)
ans  = Mul(ans, a, mod);
a = Mul(a, a, mod);
k >>= 1;
}
return ans;
}
struct Mat Mul(struct Mat a, struct Mat b, int mod){
struct Mat ans;
memset(ans.rx, 0, sizeof(ans.rx));
for(int i = 0; i < 2; i++){
for(int j = 0; j < 2; j++){
for(int k = 0; k < 2; k++){
ans.rx[i][j] += (a.rx[i][k]*b.rx[k][j])%mod;
ans.rx[i][j] %= mod;
}
}
}
return ans;
}