c/c++语言开发共享BZOJ5118: Fib数列2(二次剩余)

题意

题目链接

题目链接

一种做法是直接用欧拉降幂算出(2^p pmod{p – 1})然后矩阵快速幂。

但是今天学习了一下二次剩余，也可以用通项公式+二次剩余做。

就是我们猜想(5)在这个模数下有二次剩余，拉个板子发现真的有。

然求出来直接做就行了

#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, mod = 1125899839733759, inf = 1e9 + 10, inv2 = (mod + 1ll) >> 1ll; 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> 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; } namespace tworemain { int fmul(int a, int p, int mod = mod) {     int base = 0;     while(p) {         if(p & 1) base = (base + a) % mod;         a = (a + a) % mod; p >>= 1;     }     return base; } int fp(int a, int p, int mod = mod) {     int base = 1;     while(p) {         if(p & 1) base = fmul(base, a, mod);         p >>= 1; a = fmul(a, a, mod);     }     return base; } int f(int x) {     return fp(x, (mod - 1) >> 1); } struct mycomplex {     int a, b;     mutable int cn;     mycomplex operator * (const mycomplex &rhs)  {         return {             add(fmul(a, rhs.a, mod), fmul(cn, fmul(b, rhs.b, mod), mod)),             add(fmul(a, rhs.b, mod), fmul(b, rhs.a, mod)),             cn         };     } }; mycomplex fp(mycomplex a, int p) {     mycomplex base = {1, 0, a.cn};     while(p) {         if(p & 1) base = base * a;         a = a * a; p >>= 1;     }     return base; } int twosqrt(int n) {     if(f(n) == mod - 1) return -1;     if(f(n) ==  0) return  0;     int a = -1, val = -1;     while(val == -1) {         a = rand() << 15 | rand();         val = add(mul(a, a), -n);         if(f(val) != mod - 1) val = -1;     }     return fp({a, 1, val}, (mod + 1) / 2).a; } } using namespace tworemain; signed main() {     int rm5 = twosqrt(5), inv5 = fp(rm5, mod - 2);     int a = fmul(add(1, rm5), inv2),             b = fmul(add(1, -rm5 + mod), inv2);     int t = read();     while(t--) {         int n = read(); int pw2 = fp(2, n, mod - 1);         int x = fp(a, pw2), y = fp(b, pw2);         cout << fmul(x - y + mod, inv5) << 'n';     }     return 0; } /* 2 2 124124 */