generator 1

题目链接

题意

计算一个递推式的第n项,n是一个大数。

思路

很容易想到矩阵快速幂,但是为了避免一个把这个大数分解成二进制的过程,我们使用十进制的快速幂。

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using namespace std;
typedef long long ll;
ll mod;
struct
{
ll a[2][2];
void init()
{
a[0][0] = 0;
a[1][0] = 0;
a[1][1] = 0;
a[0][1] = 0;
}
inline ll* operator [] (int i)
{
return a[i];
}
};

inline matrix mult(matrix a, matrix b)
{
matrix ans;
ans.init();
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++)
ans[i][j] = (ans[i][j] + a[i][k] * b[k][j] % mod) % mod;
return ans;
}
char s[1000005];

inline matrix Pow_q(matrix a, ll b)
{
matrix ans;
ans.init();
ans[0][0] = 1;
ans[1][1] = 1;
while (b)
{
if (b & 1)
ans = mult(ans, a);
a = mult(a, a);
b >>= 1;
}
return ans;
}

matrix Pow(matrix a)
{
int len = strlen(s);
matrix ans;
ans.init();
ans[0][0] = 1;
ans[1][1] = 1;
for (int i = len - 1; i >= 0; i--)
{
if (s[i] != '0')
{
//dbg(s[i] - '0');
ans = mult(ans, Pow_q(a, s[i] - '0'));
}
a = Pow_q(a, 10);
}
//dbg(ans[0][0], ans[0][1], ans[1][0], ans[1][1]);
return ans;
}
*/
matrix base[15];
int main()
{
ll x0, x1, a, b;
scanf("%lld%lld%lld%lld", &x0, &x1, &a, &b);
ll n;
scanf("%s%lld", s, &mod);
int len = strlen(s);
if (len == 1 && s[0] == '0')
{
printf("%lldn", x0);
return 0;
}
if (len == 1 && s[0] == '1')
{
printf("%lldn", x1);
return 0;
}
s[len - 1]--;
int i;
for (i = len - 1; i >= 0; i--)
if (s[i] < '0')
s[i] = '9', s[i - 1]--;
else
break;
if (s[0] == '0')
{
for (int i = 0; i < len - 1; i++)
s[i] = s[i + 1];
s[len - 1] = 0;
}
//puts(s);
matrix lcy;
lcy[0][0] = a;
lcy[0][1] = b;
lcy[1][0] = 1ll;
lcy[1][1] = 0;
len = strlen(s);
matrix ans;
ans.init();
ans[0][0] = 1;
ans[1][1] = 1;
base[0].init();
base[0][0][0] = 1;
base[0][1][1] = 1;
for (int i = 1; i <= 9; i++)
base[i] = mult(base[i - 1], lcy);
for (int i = 0; i < len; i++)
{
ans = Pow_q(ans, 10);
if (s[i] != '0')
{
//dbg(s[i] - '0');
ans = mult(ans, base[s[i] - '0']);
}
}
ll ret = x1 * ans[0][0] % mod + x0 * ans[0][1] % mod;
ret %= mod;
printf("%lldn", ret);
return 0;
}