中国剩余定理及其拓展

中国剩余定理

实质就是解nnn次互质的方程，然后分别乘以他们的取模剩余量，然后相加得到答案，这里就不展开叙述。

typedef long long ll;
const int N = 1e3 + 10;
int a[N], b[N], n;
void exgcd(ll a, ll b, ll &x, ll &y) {if(!b) {x = 1;y = 0;return ;}exgcd(b, a % b, x, y);ll temp = x;x = y;y = temp - a / b * y;return ;
}
void solve() {ll m = 1, x, y, ans = 0;for(int i = 1; i <= n; i++)m *= b[i];for(int i = 1; i <= n; i++) {ll t = m / b[i];exgcd(t, b[i], x, y);ans = (ans + x * t * a[i]) % m;}ans = (ans + m) % m;printf("%lld\n", ans);
}
int main() {scanf("%d", &n);for(int i  = 1; i <= n; i++)scanf("%d %d", &a[i], &b[i]);//a[i]是余数，b[i]是模数。solve();return 0;
}

中国剩余定理拓展

思路

我们规定b[i]b[i]b[i]时余数，a[i]a[i]a[i]是模数。

显然对于一个方程我们可以直接得到ans1=b[1]ans_1 = b[1]ans1=b[1]，用这个答案去求解第二个方程，解显然可以描述为ans2=ans1+k∗a[1]ans_2 = ans_1 + k * a[1]ans2=ans1+k∗a[1]，所以第二个方程的解可以如下推理:

ans1+k∗a[1]≡b[2](moda[2])ans1 + k * a[1] \equiv b[2] \pmod {a[2]}ans1+k∗a[1]≡b[2](moda[2])

k∗a[1]≡b[2]−ans1(moda[2])k * a[1] \equiv {b[2] - ans1} \pmod{a[2]}k∗a[1]≡b[2]−ans1(moda[2])

所以我们只要解的kkk就行，无非就是做一次exgcdexgcdexgcd嘛，得到

ans2=ans1+k∗a[1]ans_2 = ans_1 + k * a[1]ans2=ans1+k∗a[1]

如果继续拓展下去，我们显然有ansk=i=1i=k−1lcm∗k+ansk−1ans_k = _{i = 1} ^ {i = k - 1} lcm * k + ans_{k - 1}ansk=i=1i=k−1lcm∗k+ansk−1，也就是把上面推导的式子换成前k−1k - 1k−1项的lcmlcmlcm。

P4777 【模板】扩展中国剩余定理（EXCRT）

裸题。

/*Author : lifehappy
*/
// #pragma GCC optimize(2)
// #pragma GCC optimize(3)
// #include <bits/stdc++.h>#include <cstdio>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
#include <cmath>#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 1e5 + 10;ll a[N], b[N];
int n;ll quick_mult(ll a, ll b, ll mod) {ll ans = 0;while(b) {if(b & 1) ans = (ans + a) % mod;b >>= 1;a = (a + a) % mod;}return ans;
}ll ex_gcd(ll a, ll b, ll & x, ll & y) {if(!b) {x = 1, y = 0;return a;}ll gcd = ex_gcd(b, a % b, x, y);ll temp = x;x = y;y = temp - a / b * y;return gcd;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read();for(int i = 1; i <= n; i++) a[i] = read(), b[i] = read();ll ans = b[1], M = a[1];for(int i = 2; i <= n; i++) {ll A = M, B = a[i], C = ((b[i] - ans) % a[i] + a[i]) % a[i], x, y;ll gcd = ex_gcd(A, B, x, y);B /= gcd;x = quick_mult(x, C / gcd, B);ll mod = M * B;ans = (ans + quick_mult(x, M, mod)) % mod;M = mod;}printf("%lld\n", ((ans % M) + M) % M);return 0;
}