洛谷题单指南-进阶数论-P1495 【模板】中国剩余定理（CRT）/ 曹冲养猪

原题链接：https://www.luogu.com.cn/problem/P1495

题意解读：求方程组x ≡ b_i (mod a_i), i∈[1,n]的最小正整数解，所有的a_i互质。

解题思路：

1、中国剩余定理

设方程组为（a1，a2，a3互质）：

• x ≡ b1 (mod a1)
• x ≡ b2 (mod a2)
• x ≡ b3 (mod a3)

要找到x满足上面三种情况，设a = a1 * a2 * a3

x需要拆解成三个数：x = (x1 + x2 + x3) % a，

必须满足：

• x1 % a1 = b1, x1 % a2 = 0, x1 % a3 = 0
• x2 % a2 = b2, x2 % a1 = 0, x2 % a3 = 0
• x3 % a3 = b3, x3 % a1 = 0, x3 % a2 = 0

可以看出，

x1是a2、a3的倍数，不妨设x1 = a2 * a3 * k1，a2 * a3 * k1 ≡ b1 (mod a1)，

转化为不定方程a2 * a3 * k1 + a1 * p1 = b1，

由于gcd(a2 * a3, a1) = 1，求a2 * a3 * k1 + a1 * p1 = 1的一个k1的解再乘以b1倍即可

又有a2 * a3 * k1 ≡ 1 (mod a1)，k1的解就是a2 * a3模a1的逆元，再乘以b1

因此：

k1 = (a2 * a3)^-1 * b1

x1 = a2 * a3 * (a2 * a3)^-1 * b1

同理分析，可得到：

x2 = a1 * a3 * (a1 * a3)^-1 * b2

x3 = a1 * a2 * (a1 * a2)^-1 * b3

x = a2 * a3 * (a2 * a3)^-1 * b1 + a1 * a3 * (a1 * a3)^-1 * b2 + a1 * a2 * (a1 * a2)^-1 * b3

= (a / a1) * (a / a1)^-1 * b1 + (a / a2) * (a / a2)^-1 * b2 + (a / a3) * (a / a3)^-1 * b3

x最后要模a

推而广之：

设a = a1 * a2 * ... * an

ti = a / ai，ti^-1是ti在模ai意义下的逆元

x = (∑ ti * ti^-1 * bi) % a

100分代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;const int N = 15;
LL a[N], b[N];
LL n, A = 1, ans;LL exgcd(LL a, LL b, LL &x, LL &y)
{if(b == 0){x = 1, y = 0;return a;}LL d = exgcd(b, a % b, y, x);y = y - a / b * x;return d;
}LL mul(LL a, LL b, LL mod)
{LL res = 0;while(b){if(b & 1) res = (res + a) % mod;b >>= 1;a = (a + a) % mod;}return res;
}int main()
{cin >> n;for(int i = 1; i <= n; i++){cin >> a[i] >> b[i];A *= a[i];}for(int i = 1; i <= n; i++){LL m = A / a[i];LL x, y;exgcd(m, a[i], x, y); //求m模a[i]的逆元x = (x % a[i] + a[i]) % a[i];//ans = (ans + m * x * b[i]) % A; //可能溢出ans = (ans + mul(m, x * b[i], A)) % A; //快速乘避免溢出}cout << ans;return 0;
}

2、扩展：同时适用于模数不互质的情况

设方程组：

• x ≡ b1 (mod a1)
• x ≡ b2 (mod a2)
• ...
• x ≡ bn (mod an)

先取前两个方程，展开

x = a1 * s + b1

x = a2 * t + b2

a1 * s - a2 * t = b2 - b1

调整负数为整数，因为t是任意值，所以系数正负没关系，正数便于计算

a1 * s + a2 * t = b2 - b1

用扩展欧几里得算法求a1 * s + a2 * t = gcd(a1, a2)的一个特解s0

如果(b2 - b1) % d != 0，则方程组无解！

设d = gcd(a1, a2), tmp1 = (b2 - b1) / d，tmp2 = a2 / d

s的通解为：s = s0 * tmp1 + tmp2 * k，k为任意整数 

注意：在计算s0 * tmp1时如果数据范围过大可考虑模tmp2的快速乘！前提是tmp1是非负数，即tmp1 = (tmp1 % tmp2 + tmp2) % tmp2

将s的通解代入x = a1 * s + b1得到x = a1 * (s0 * tmp1 + tmp2 * k) + b1，展开得到

x = a1 * tmp2 * k + (a1 * s0 * tmp1 + b1) 

将方程x = a1 * tmp2 * k + (a1 * s0 * tmp1 + b1) 看作x = a1 * s + b1，可以将a1 = a1 * tmp2, b1 = a1 * s0 * tmp1 + b1

将方程x = a3 * s + b3看作x = a2 * s + b2

继续上述合并方程求s的过程，最后处理完所有的方程，对(b1 % a1 + a1) % a1即得到最小正整数解。

100分代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;const int N = 15;
LL a[N], b[N];
LL n, A = 1, ans;LL exgcd(LL a, LL b, LL &x, LL &y)
{if(b == 0){x = 1, y = 0;return a;}LL d = exgcd(b, a % b, y, x);y = y - a / b * x;return d;
}LL mul(LL a, LL b, LL mod)
{LL res = 0;while(b){if(b & 1) res = (res + a) % mod;b >>= 1;a = (a + a) % mod;}return res;
}int main()
{cin >> n;for(int i = 1; i <= n; i++){cin >> a[i] >> b[i];}LL a1 = a[1], b1 = b[1];for(int i = 2; i <= n; i++){int a2 = a[i], b2 = b[i];LL s, t;LL d = exgcd(a1, a2, s, t);if((b2 - b1) % d){//无解的情况此题不用考虑}LL tmp1 = (b2 - b1) / d, tmp2 = a2 / d;tmp1 = (tmp1 % tmp2 + tmp2) % tmp2;s = mul(s, tmp1, tmp2);b1 = a1 * s + b1;a1 = a1 * tmp2;}cout << (b1 % a1 + a1) % a1;return 0;
}