B:generator 1
題意
給你讓你求出
思路
典型的矩陣快速冪,但是n的範圍太大,所以得快速冪得用十進制快速冪
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 1e6 + 5;
ll mod;
struct Matrix{
ll mat[2][2];
Matrix() {memset(mat, 0, sizeof(mat));};
void init() {
mat[0][0] = mat[1][1] = 1;
}
void init(ll a, ll b) {
mat[0][0] = 0; mat[0][1] = b;
mat[1][0] = 1; mat[1][1] = a;
}
void operator = (Matrix x) {
for (int i = 0; i <= 1; i ++)
for (int j = 0; j <= 1; j ++)
mat[i][j] = x.mat[i][j];
}
};
void Print(Matrix x) {
for (int i = 0; i <= 1; i ++) {
for (int j = 0; j <= 1; j ++)
cout << x.mat[i][j] << " ";
cout << endl;
}
}
Matrix operator * (Matrix x, Matrix y) {
Matrix t;
for (int i = 0; i <= 1; i ++)
for (int j = 0; j <= 1; j ++)
for (int k = 0; k <= 1; k ++)
t.mat[i][j] = (t.mat[i][j] + x.mat[i][k] * y.mat[k][j]) % mod;
return t;
}
Matrix Ksm(Matrix x, ll b) {
//cout << b << endl;
Matrix t; t.init();
while(b) {
if(b & 1) t = t * x;
x = x * x;
b >>= 1;
}
//Print(t);
return t;
}
int main() {
ll x0, x1, a, b;
scanf("%lld %lld %lld %lld", &x0, &x1, &a, &b);
char s[maxn];
scanf("%s%lld", s, &mod);
int len = strlen(s);
reverse(s, s+len);
Matrix t, ans; t.init(a, b);
ans.mat[0][0] = x0; ans.mat[0][1] = x1;
Matrix res;
res.init();
for (int i = 0; i < len; i ++) {
res = res * Ksm(t, s[i]-'0');
t = Ksm(t, 10);
// Print(res);
// Print(t);
}
ans = ans * res;
printf("%lld\n", ans.mat[0][0]);
return 0;
}
C:generator 2
題意
有這麼一個遞推式,讓你求在中第一次出現的位置
思路
因爲遞推式模,所以的循環節一定小於,
而又是這種形式
所以我們的任務就變成求最小的
而明顯可以用BSGS
但是BSGS的一個使用條件能不能求出
但是我們怎麼求出呢
正常的加是乘a加b,那麼除就是除a減
舉個例子:
我們從降到,先除再減去變成,然後再除減去變成
那我們從降到只需要進行次操作即可
這樣我們就可以用BSGS了
跟BSGS的步驟差不多,我們可以把式子化成
我們可以預處理出來 ,然後遍歷找到一個
也就是說在這個式子中不是一個值,而是一種操作,把所代表的次數降下
因爲我們已經預處理了一個,那麼在我們遍歷的過程中每次降下一個,知道找到或者找不到
用Hash存一下
AC代碼
#include<bits/stdc++.h>
using namespace std;
#define ll long long
typedef pair<int, int>pis;
const int limit = 1e6;
pis d[limit+6];
int vals[limit+6], pos[limit+6];
int Ksm(ll a, int b, int p) {
ll res = 1;
while(b) {
if(b & 1) res = res * a % p;
a = a * a % p;
b >>= 1;
}
return res;
}
int inv(int a, int p) { return Ksm(a, p-2, p); }
void solve() {
ll n, x0, a, b, p; int Q;
scanf("%lld %lld %lld %lld %lld %d", &n, &x0, &a, &b, &p, &Q);
if(!a) {
while(Q --) {
int v; scanf("%d", &v);
if(v == x0) printf("0\n");
else if(v == b) printf("1\n");
else printf("-1\n");
}
return ;
}
d[0] = {x0, 0};
for (int i = 1; i <= limit; i ++) {
int val = (a*d[i-1].first+b) % p;
d[i] = {val, i};
}
sort(d, d+limit+1);
int cnt = 0;
for (int i = 0; i <= limit; i ++) {
vals[cnt] = d[i].first; pos[cnt++] = d[i].second;
while(d[i].first == d[i+1].first && i+1 <= limit) i++;
}
int inv_a = inv(a, p);
int inv_b = (p-b) % p * inv_a % p;
ll aa = 1, bb = 0;
for (int i = 0; i <= limit; i ++) {
aa = aa * inv_a % p;
bb = (bb * inv_a + inv_b) % p;
}
while(Q --) {
int v; scanf("%d", &v);
int it = lower_bound(vals, vals+cnt, v) - vals;
if(it < cnt && vals[it] == v) {
if(pos[it] < n) printf("%d\n", pos[it]);
else printf("-1\n");
continue;
}
int m = p/(limit+1) + 3, flag = 0;
for (int i = 1; i <= m; i ++) {
v = (aa * v + bb) % p;
it = lower_bound(vals, vals+cnt, v) - vals;
if(it<cnt && vals[it] == v) {
flag = 1;
int res = i*(limit+1)+pos[it];
if(res>=n) res = -1;
printf("%d\n", res);
break;
}
}
if(!flag) printf("-1\n");
}
}
int main() {
int T;
scanf("%d", &T);
while(T --) solve();
return 0;
}