【洛谷P1495】曹冲养猪

Description

给定若干个形如$x\equiv a_i \pmod {m_i}$的同余方程，其中m两两互质，求x

Solution

这是中国剩余定理的模板题，具体解法如下：

我们首先求出$M=\prod\limits_{i=1}^{n}{m_i}$

那么令$M_i=\frac{M}{m_i}$

然后令$t_i=M_i^{-1} \pmod {m_i}$

最后令$s_i=t_iM_i$

我们只要求出$x\equiv \sum\limits_{i=1}^{n}{a_is_i}\pmod M$即可

逆元可以用Exgcd求解，时间复杂度为O(nlogn)

Code

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 inline int read() {
 5     int ret = 0, op = 1;
 6     char c = getchar();
 7     while (!isdigit(c)) {
 8         if (c == '-') op = -1; 
 9         c = getchar();
10     }
11     while (isdigit(c)) {
12         ret = ret * 10 + c - '0';
13         c = getchar();
14     }
15     return ret * op;
16 }
17 ll exgcd(ll a, ll b, ll &x, ll &y) {
18     if (!b) {
19         x = 1, y = 0;
20         return a;
21     }
22     ll gcd = exgcd(b, a % b, x, y);
23     ll x2 = x, y2 = y;
24     x = y2;
25     y = x2 - (a / b) * y2;
26     return gcd;
27 }
28 ll n, a[12], b[12], mod = 1;
29 ll ans;
30 int main() {
31     n = read();
32     for (register int i = 1; i <= n; ++i) {
33         a[i] = read(), b[i] = read();
34         mod *= a[i];
35     }
36     for (register int i = 1; i <= n; ++i) {
37         ll m = mod / a[i];
38         ll x = 0, y = 0;
39         exgcd(m, a[i], x, y);
40         ans = (ans + (x * m * b[i]) % mod + mod) % mod;
41     }
42     printf("%lld\n", ans);
43     return 0;
44 }

AC Code