【SDOI2011】计算器

Description

给定$y, z, p$，求$x=y^{z} \mod p$或$xy\equiv z \pmod p$或$y^x\equiv z\pmod p$中x的值

Solution

第一个式子我们可以直接用快速幂求解答案。时间复杂度$O(log_{2}z)$

第二个式子我们可以变形为$xy+Yp=z$，然后用扩展欧几里得算法求出$xy+Yp=gcd(y,p)$的解，然后利用裴蜀定理判断是否有解即可。

第三个式子我们可以应用BSGS算法求解。详细的解释在这里

Code

 1 #include <cstdio>
 2 #include <map>
 3 #include <cmath>
 4 using namespace std;
 5 typedef long long ll;
 6 ll qpow(ll a, ll b, ll p) {
 7     ll ret = 1;
 8     while (b) {
 9         if (b & 1) ret = (ret * a) % p;
10         a = a * a % p;
11         b >>= 1;
12     }
13     return ret % p;
14 }
15 ll exgcd(ll a, ll b, ll &x, ll &y) {
16     if (!b) {
17         x = 1, y = 0;
18         return a;
19     }
20     ll gcd = exgcd(b, a % b, x, y);
21     ll x2 = x, y2 = y;
22     x = y2;
23     y = x2 - (a / b) * y2;
24     return gcd;
25 }
26 map <ll, ll> h;
27 ll BSGS(ll a, ll b, ll p) {
28     h.clear();
29     b %= p;
30     ll t = (ll)sqrt(p) + 1;
31     for (register int j = 0; j < t; ++j) {
32         ll val = (ll)b * qpow(a, j, p) % p;
33         h[val] = j;
34     }
35     a = qpow(a, t, p);
36     if (a == 0) return b == 0 ? 1 : -1;
37     for (register int i = 0; i <= t; ++i) {
38         ll val = qpow(a, i, p);
39         ll j = h.find(val) == h.end() ? -1 : h[val];
40         if (j >= 0 && i * t - j >= 0) return i * t - j;
41     }
42     return -1;
43 }
44 int n, k;
45 int main() {
46     ll y = 0, z = 0, p = 0;
47     scanf("%d%d", &n, &k);
48     if (k == 1) 
49         while (n--) {
50             scanf("%lld%lld%lld", &y, &z, &p);
51             printf("%lld\n", qpow(y, z, p));
52         }
53     else if (k == 2) 
54         while (n--) {
55             scanf("%lld%lld%lld", &y, &z, &p);
56             ll x = 0, yy = 0;
57             ll ret = exgcd(y, p, x, yy);
58             if (z % ret != 0) {puts("Orz, I cannot find x!"); continue ;}
59             ll now = p / ret;
60             while (x < 0) x += now;
61             printf("%lld\n", (x * (z / ret) % now) + now % now);
62         }
63     else 
64         while (n--) {
65             scanf("%lld%lld%lld", &y, &z, &p);
66             ll ret = BSGS(y, z, p);
67             if (ret == -1) puts("Orz, I cannot find x!");
68             else printf("%lld\n", ret);
69         }
70     return 0;
71 }

AC Code

 

标签：return,ll,ret,SDOI2011,lld%,计算器,now,lld
来源： https://www.cnblogs.com/shl-blog/p/11332183.html