二次剩余

参考资料。

类似fft，自定义一种复数，使得第二维的平方等于w。

然后就可以进行SAO操作了。

注意sgn要返回long long

例题：Timus Online Judge 1132

 1 #include 
 2 #include 
 3 #include 
 4 
 5 typedef long long LL;
 6 const int N = 100010;
 7 
 8 LL w, MO;
 9 
10 inline int rd() {
11     return (rand() * (1ll << 16) + rand()) % LONG_MAX;
12 }
13 
14 struct cp {
15     LL x, y;
16     cp(LL X = 0, LL Y = 0) {
17         x = X;
18         y = Y;
19     }
20     inline cp operator *(const cp &W) const {
21         return cp(x * W.x % MO + y * W.y % MO * w % MO, x * W.y % MO + y * W.x % MO);
22     }
23     inline cp operator %(const LL &W) const {
24         return cp(x % W, y % W);
25     }
26 };
27 
28 template <class T> inline T qpow(T a, LL b) {
29     T ans(1);
30     while(b) {
31         if(b & 1) ans = ans * a % MO;
32         a = a * a % MO;
33         b = b >> 1;
34     }
35     return ans;
36 }
37 
38 inline int sgn(LL x) {
39     return qpow(x, (MO - 1) >> 1);
40 }
41 
42 inline void solve() {
43     LL n;
44     scanf("%lld%lld", &n, &MO);
45     n %= MO;
46     if(MO == 2) {
47         printf("%d\n", n);
48         return;
49     }
50     if(sgn(n) == MO - 1) {
51         printf("No root\n");
52         return;
53     }
54     LL a;
55     while(1) {
56         a = rd() % MO;
57         w = (a * a % MO - n + MO) % MO;
58         if(sgn(w) == MO - 1) break;
59     }
60     cp ans(a, 1);
61     ans = qpow(ans, (MO + 1) >> 1);
62     LL b = ans.x, c = MO - b;
63     if(b > c) std::swap(b, c);
64     printf("%lld %lld \n", b, c);
65     return;
66 }
67 
68 int main() {
69     srand(19384);
70     int T;
71     scanf("%d", &T);
72     while(T--) solve();
73     return 0;
74 }

AC代码

 

转载于:https://www.cnblogs.com/huyufeifei/p/10560629.html