2019牛客暑期多校训练营（第五场）B generator 1 十进制快速幂 / 循环节 积性

链接：https://ac.nowcoder.com/acm/contest/885/B?&headNav=acm&headNav=acm&headNav=acm
来源：牛客网
 

generator 1

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

You are given four positive integers x0,x1,a,bx_0, x_1, a, bx0​,x1​,a,b. And you know xi=a⋅xi−1+b⋅xi−2x_i = a \cdot x_{i-1} + b \cdot x_{i-2}xi​=a⋅xi−1​+b⋅xi−2​ for all i≥2i \ge 2i≥2.

Given two positive integers n, and MOD, please calculate xnx_nxn​ modulo MOD.

Does the problem look simple? Surprise! The value of n may have many many digits!

输入描述:

The input contains two lines.
The first line contains four integers x0,x1,a,bx_0, x_1, a, bx0​,x1​,a,b (1≤x0,x1,a,b≤1091 \le x_0, x_1, a, b \le 10^91≤x0​,x1​,a,b≤109).
The second line contains two integers n, MOD (1≤n<10(106),109 
  
输出描述:
 
  
Print one integer representing the answer.
 
  
示例1
 
  
输入
 
  
复制
 
  
1 1 1 1
10 1000000001
 
  
输出
 
  
复制
 
  
89
 
  
说明
 
  
The resulting sequence x is Fibonacci sequence. The 11-th item is 89.
 
  
示例2
 
  
输入
 
  
复制
 
  
1315 521 20185 5452831
9999999999999999999999999999999999999 1000000007
 
  
输出
 
  
复制
 
  
914730061
 
  
题解1：打表可以发现，mod数的结果存在循环节，并且符合积性函数的性质，即：
 
  
1、f(a ) = (a - 1)*(a +1)(a为质数)
 
  
2、f(a*b) = f(a) * f(b) (a , b 互质)
 
  
3、f(a*b) = f(a) * b (a%b==0)
 
  
所以，先求出循环节，然后快速幂，注意要用到快速乘
 
  
#include 
using namespace std;
 
typedef long long ll;
const int N = 1e6 + 10;
struct node {
    ll mat[2][2];
};
  
int prime[N], len;
int ok[N];
ll x0, x1, a, b, mod, p;
node ans, O, tmp, z;
char n[N];
 
void init() {
    for(int i = 2; i < N; i++) {
        if(!ok[i])
            prime[len++] = i;
        for(int j = 0; j < len && (ll)i * prime[j] < N; j++) {
            ok[i * prime[j]] = 1;
            if(i % prime[j] == 0) break;
        }
    }
     
}
 
ll ksc (ll x, ll y, ll z) {
    x %= z;
    y %= z;
    ll res = 0;
    while(y) {
        if(y & 1)
            res = (res + x) % z;
        x = (x + x) % z;
        y >>= 1;
    }
    return res;
}
 
node cul (node x, node y) {
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++) {
            z.mat[i][j] = 0;
            for(int k = 0 ; k < 2; k++) {
                z.mat[i][j] = (z.mat[i][j] + ksc(x.mat[i][k] , y.mat[k][j], mod)) % mod;
            }
        }
    return z;
}
 
ll getmod(ll x) {
    ll ans1 = 1, ans2 = 1, cnt = x;
    for(int i = 0; i < len && (ll)prime[i] * prime[i] <= x; i++) {
        if(x % prime[i] == 0){
            ans1 *= (prime[i] - 1) * (prime[i] + 1);
            ans2 *= prime[i];
            while(x % prime[i] == 0) x /= prime[i];
        }
    }
    if(x > 1) {
        ans1 *= (x - 1) * (x + 1);
        ans2 *= x;
    }
 //   cout << cnt << " " << ans1 << " " << ans2 << endl;
    return cnt / ans2 * ans1;
}
 
void solve(ll x) {
    ans.mat[0][0] = ans.mat[1][1] = 1;
    ans.mat[0][1] = ans.mat[1][0] = 0;
    tmp.mat[0][0] = a;
    tmp.mat[0][1] = b;
    tmp.mat[1][0] = 1;
    tmp.mat[1][1] = 0;
    while(x) {
        if(x & 1) ans = cul(ans, tmp);
        tmp = cul(tmp, tmp);
        x >>= 1;
    }
     
    printf("%lld\n", (ksc(ans.mat[1][0], x1, mod) + ksc(ans.mat[1][1], x0, mod)) % mod);
}
 
int main() {
     
    init();
    ll x = 0;
    scanf("%lld %lld %lld %lld %s %lld", &x0, &x1, &a, &b, n, &mod);
    int len = strlen(n);
    p = getmod(mod);
 //   cout << mod << endl;
    for(int i = 0; i < len; i++) {
        x = ksc (x , 10, p) ;
        x = (x + n[i] -'0') % p;
    }
    solve(x);
    return 0;
}
 
  
题解2：十倍增的快速幂
 
  
#include 
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
struct node {
    ll mat[2][2];
};
ll x0, x1, a, b, mod;
node ans, O, tmp, z;
char n[N];
node cul (node x, node y) {
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++) {
            z.mat[i][j] = 0;
            for(int k = 0 ; k < 2; k++)
                z.mat[i][j] = (z.mat[i][j] + x.mat[i][k] * y.mat[k][j] ) % mod;
        }
    return z;
}
int main() {
    int op;
    scanf("%lld %lld %lld %lld %s %lld", &x0, &x1, &a, &b, n, &mod);
    int len = strlen(n);
    ans.mat[0][0] = ans.mat[1][1] = 1;
    tmp.mat[0][0] = a;
    tmp.mat[0][1] = b;
    tmp.mat[1][0] = 1;
    for(int i = len - 1; i >= 0; i--) {
        op = n[i] - '0';
        for(int j = 0; j < op; j++) ans = cul(ans, tmp);
        O.mat[0][0] = O.mat[1][1] = 1;
        O.mat[0][1] = O.mat[1][0] = 0;
        for(int j = 0; j < 10; j++) O = cul(O, tmp);
        tmp = O;
    }
    printf("%lld\n", (x1 * ans.mat[1][0] + x0 * ans.mat[1][1] ) % mod) ;
    return 0;
}