CRT && exCRT模板

CRT从各种方面上都吊打exCRT啊......

短，好理解...

考虑构造bi使得bi % pi = ai，bi % pj = 0。然后全加起来就行了。

显然bi的构造就是ai * (P/pi) * inv(P/pi)。

LL a = 0, p = MO - 1;
for(int i = 1; i <= 4; i++) {
    a = (a + ans[i] * (p / mod[i]) % p * qpow(p / mod[i], mod[i] - 2, mod[i]) % p) % p;
}

exCRT：

是这样的，重新手推了一个短一点的模板。题是洛谷P3868 猜数字

 1 inline int exCRT(int n, int *a, int *b) {
 2     int t = a[1], p = b[1], x, y;
 3     for(int i = 2; i <= n; i++) {
 4         int g = exgcd(p, b[i], x, y);
 5         p = lcm(p, b[i]);
 6         t = (t - a[i]) % p;
 7         y = y * (t / g) % p;
 8         t = (a[i] + y * b[i]) % p;
 9     }
10     return t;
11 }

具体操作的时候开long long，龟速乘，记得全程避免负数。

先背为敬。

 1 #include <cstdio>
 2 #include <algorithm>
 3 
 4 typedef long long LL;
 5 const int N = 100010;
 6 
 7 LL p[N], a[N];
 8 
 9 inline LL mod(LL a, LL c) {
10     if(c < 0) {
11         c = (~c) + 1;
12     }
13     while(a >= c) {
14         a -= c;
15     }
16     while(a < 0) {
17         a += c;
18     }
19     return a;
20 }
21 inline LL mul(LL a, LL b, LL c) {
22     LL ans = 0;
23     while(b) {
24         if(b & 1) {
25             ans = mod(ans + a, c);
26         }
27         a = mod(a << 1, c);
28         b = b >> 1;
29     }
30     return ans;
31 }
32 LL exgcd(LL a, LL b, LL &x, LL &y) {
33     if(!b) {
34         x = 1;
35         y = 0;
36         return a;
37     }
38     LL g = exgcd(b, a % b, x, y);
39     std::swap(x, y);
40     y -= (a / b) * x;
41     return g;
42 }
43 
44 int main() {
45     int n;
46     scanf("%d", &n);
47     for(int i = 1; i <= n; i++) {
48         scanf("%lld%lld", &p[i], &a[i]);
49     }
50 
51     LL A = a[1], P = p[1];
52     for(int i = 2; i <= n; i++) {
53         LL x, y;
54         LL C = (a[i] - A), g = exgcd(P, p[i], x, y);
55         C = (C % p[i] + p[i]) % p[i];
56         if(C % g) {
57             puts("-1");
58             return 0;
59         }
60         
61         x = mul(x, C / g, P / g * p[i]);
62         A += mul(x, P, P / g * p[i]);
63         P *= p[i] / g;
64         A = mod(A, P);
65     }
66     
67     // x === A mod P
68     LL x, y;
69     exgcd(P, 1, y, x);
70     x *= A;
71     x = (x % P + P) % P;
72     printf("%lld\n", x);
73     return 0;
74 }