中国剩余定理

  1 //author Eterna
  2 #define Hello the_cruel_world!
  3 #pragma GCC optimize(2)
  4 #include<iostream>
  5 #include<algorithm>
  6 #include<cstdio>
  7 #include<string>
  8 #include<cstring>
  9 #include<vector>
 10 #include<map>
 11 #include<set>
 12 #include<queue>
 13 #include<stack>
 14 #include<utility>
 15 #include<cmath>
 16 #include<climits>
 17 #include<deque>
 18 #include<functional>
 19 #include<complex>
 20 #include<numeric>
 21 #include<unordered_map>
 22 #define max(x,y) ((x)>(y)?(x):(y))
 23 #define min(x,y) ((x)<(y)?(x):(y))
 24 #define Pi acos(-1.0)
 25 #define ABS(x) ((x) >= 0 ? (x) : (-(x)))
 26 #define pb(x) push_back(x)
 27 #define lowbit(x) (x & -x)
 28 #define FRIN freopen("C:\\Users\\Administrator.MACHENI-KA32LTP\\Desktop\\in.txt", "r", stdin)
 29 #define FROUT freopen("C:\\Users\\Administrator.MACHENI-KA32LTP\\Desktop\\out.txt", "w", stdout)
 30 #define FAST ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
 31 #define outd(x) printf("%d\n", x)
 32 #define outld(x) printf("lld\n", x)
 33 #define il inline
 34 using namespace std;
 35 typedef long long ll;
 36 typedef unsigned long long ull;
 37 typedef pair<int, int> pii;
 38 const int maxn = 100;
 39 const int INF = 0x7fffffff;
 40 const int mod = 1e9 + 7;
 41 const double eps = 1e-7;
 42 inline int read_int() {
 43     char c;
 44     int ret = 0, sgn = 1;
 45     do { c = getchar(); } while ((c < ‘0‘ || c > ‘9‘) && c != ‘-‘);
 46     if (c == ‘-‘) sgn = -1; else ret = c - ‘0‘;
 47     while ((c = getchar()) >= ‘0‘ && c <= ‘9‘) ret = ret * 10 + (c - ‘0‘);
 48     return sgn * ret;
 49 }
 50 inline ll read_ll() {
 51     char c;
 52     ll ret = 0, sgn = 1;
 53     do { c = getchar(); } while ((c < ‘0‘ || c > ‘9‘) && c != ‘-‘);
 54     if (c == ‘-‘) sgn = -1; else ret = c - ‘0‘;
 55     while ((c = getchar()) >= ‘0‘ && c <= ‘9‘) ret = ret * 10 + (c - ‘0‘);
 56     return sgn * ret;
 57 }
 58 ll a[maxn + 5], b[maxn + 5], d, res;
 59 ll gcd(ll a, ll b) {
 60     return b ? gcd(b, a % b) : a;
 61 }
 62 ll lcm(ll a, ll b) {
 63     return a / gcd(a, b) * b;
 64 }
 65 void ex_gcd(ll a, ll b, ll &x, ll &y, ll &d) {
 66     if (!b) { d = a, x = 1, y = 0; }
 67     else {
 68         ex_gcd(b, a % b, y, x, d);
 69         y -= x * (a / b);
 70     }
 71 }
 72 ll inv(ll t, ll p) {//如果不存在，返回-1 
 73     ll d, x, y;
 74     ex_gcd(t, p, x, y, d);
 75     return d == 1 ? (x % p + p) % p : -1;
 76 }
 77 ll Quick_mul(ll a, ll b, ll mod) {
 78     ll res = 0;
 79     while (b) {
 80         if (b & 1)res = (res + a) % mod;
 81         a = 2 * a % mod;
 82         b >>= 1;
 83     }
 84     return res;
 85 }
 86 int n;
 87 int main()
 88 {
 89     n = read_int();
 90     for (int i = 1; i <= n; ++i)a[i] = read_int();
 91     for (int i = 1; i <= n; ++i)b[i] = read_int();
 92     for (int i = 1; i <= n; ++i)a[i] = (a[i] % b[i] + b[i]) % b[i];
 93     d = b[1];
 94     for (int i = 2; i <= n; ++i)d = lcm(d, b[i]);
 95     for (int i = 1; i <= n; ++i) {
 96         ll p = d / b[i];
 97         ll g = inv(p, b[i]);
 98         res = (res + Quick_mul(Quick_mul(a[i], p, d), g, d)) % d;
 99     }
100     res = (res % d + d) % d;
101     cout << res << endl;
102     //system("pause");
103     return 0;
104 }