c/c++语言开发共享BZOJ3122: [Sdoi2013]随机数生成器(BSGS)

题意

题目链接

sol

这题也比较休闲。

直接把(x_{i+1} = (ax_i + b) pmod p)展开，推到最后会得到这么个玩意儿

[ a^{i-1} (x_1 + frac{b}{a-1}) – frac{b}{a-1} equiv t pmod p ]

然后再合并一下就可以大力bsgs了。

有些细节需要特判一下

#include<bits/stdc++.h>  #define pair pair<int, int> #define mp(x, y) make_pair(x, y) #define fi first #define se second #define int long long  #define ll long long  #define ull unsigned long long  #define fin(x) {freopen(#x".in","r",stdin);} #define fout(x) {freopen(#x".out","w",stdout);} using namespace std; const int maxn = 1e6 + 10, inf = 1e9 + 10;; int mod; const double eps = 1e-9; template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;} template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;} template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;} template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);} template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;} template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;} template <typename a> inline void debug(a a){cout << a << 'n';} template <typename a> inline ll sqr(a x){return 1ll * x * x;} template <typename a, typename b> inline ll fp(a a, b p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;} template <typename a> a inv(a x) {return fp(x, mod - 2);} inline int read() {     char c = getchar(); int x = 0, f = 1;     while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();     return x * f; } int a, b, x1, end; //x_{i+1} = (ax_i + b) % p //a^ans = x % p //a^{i * k - j} = x % p //a^{i * k} = x * a^j % p map<int, int> mp;  /* int query(int a, int x, int p) {     if(__gcd(a, p) != 1) return -2;     int base = 1;     for(int i = 0; i <= p; i++) {         if(base % p == x) return i;         mul2(base, a);     }     return -2; } */  int query(int a, int x, int p) {     if(__gcd(a, p) != 1) return -2;     mp.clear(); int block = ceil(sqrt(p)), base = fp(a, block);     for(int i = 0, cur = x; i <= block; i++, mul2(cur, a)) mp[cur] = i;     for(int i = 1, cur = base; i <= block; i++, mul2(cur, base))          if(mp[cur])              return i * block - mp[cur];     return -2; }  void solve() {     mod = read(); a = read(); b = read(); x1 = read(); end = read();     if(x1 == end) {puts("1"); return ;}     if(!a) {         if(!b) {puts(end == x1 ? "1" : "-1");return ;}         else {puts(end == b ? "2" : "-1");return ;}     }     if(a == 1) {         if(!b) {puts(end == x1 ? "1" : "-1");return ;}         else {             //int tmp = add(end, -x1 + mod) % b;             //cout << tmp << 'n';             cout << mul(add(end, -x1), inv(b)) + 1 << 'n';              return ;         }     }     int tmp = mul(b, inv(a - 1));     add2(x1, tmp); add2(end, tmp);      mul2(end, inv(x1));     cout << query(a, end, mod) + 1 << 'n'; } signed main() {     //freopen("a.in", "r", stdin);     for(int t = read(); t--; solve());     return 0; }