一些关于中国剩余定理的数论题(POJ 2891/HDU 3579/HDU 1573/HDU 1930)

2891 -- Strange Way to Express Integers

 import java.math.BigInteger;
 import java.util.Scanner;

 public class Main {
     static final BigInteger ZERO = new BigInteger("0");
     static final BigInteger ONE = new BigInteger("1");
     static BigInteger gcd(BigInteger a, BigInteger b) {
         if (b.equals(ZERO)) return a;
         else return gcd(b, a.mod(b));
     }
     static BigInteger lcm(BigInteger a, BigInteger b) {
         return a.divide(gcd(a, b)).multiply(b);
     }
     static void gcd(BigInteger a, BigInteger b, BigInteger p[]) {
         if (b.equals(ZERO)) {
             p[0] = a;
             p[1] = new BigInteger("1");
             p[2] = new BigInteger("0");
         } else {
             gcd(b, a.mod(b), p);
             BigInteger tmp = p[1];
             p[1] = p[2];
             p[2] = tmp;
             p[2] = p[2].subtract(a.divide(b).multiply(p[1]));
         }
     }
     static BigInteger[] A = new BigInteger[11111];
     static BigInteger[] R = new BigInteger[11111];
     static BigInteger work(int n) {
         BigInteger[] p = new BigInteger[3];
         BigInteger a = A[0], r = R[0];
         for (int i = 1; i < n; i++) {
             BigInteger dr = R[i].subtract(r);
             gcd(a, A[i], p);
             if (dr.mod(p[0]).equals(ZERO) == false) return new BigInteger("-1");
             BigInteger tmp = a.multiply(p[1]);
             a = lcm(a, A[i]);
             tmp = tmp.mod(a).add(a).mod(a);
             tmp = tmp.multiply(dr).divide(p[0]).add(r);
             r = tmp.mod(a).add(a).mod(a);
         }
         return r;
     }
     public static void main(String[] args) {
         int n;
         Scanner in = new Scanner(System.in);
         while (in.hasNext()) {
             n = in.nextInt();
             String tmp;
             for (int i = 0; i < n; i++) {
                 tmp = in.next();
                 A[i] = new BigInteger(tmp);
                 tmp = in.next();
                 R[i] = new BigInteger(tmp);
             }
             System.out.println(work(n).toString());
         }
     }
 }

　　因为懒得写大数，所以直接用java的大数类来做。用这个是因为，虽然输入输出不会超过64位整型，不过中间过程还是超了。

Problem - 3579

 #include <cstdio>
 #include <iostream>
 #include <cmath>
 #include <algorithm>
 #include <cstring>

 using namespace std;

 typedef long long LL;

 const int N = ;
 LL A[N], R[N];
 inline LL gcd(LL a, LL b) { return b ? gcd(b, a % b) : a;}
 inline LL lcm(LL a, LL b) { return a / gcd(a, b) * b;}
 void gcd(LL a, LL b, LL &d, LL &x, LL &y) {
     if (b) { gcd(b, a % b, d, y, x); y -= a / b * x;}
     else { d = a, x = , y = ;}
 }

 LL work(int n) {
     LL x, y, d, a = A[], r = R[];
     for (int i = ; i < n; i++) {
         LL dr = R[i] - r;
         gcd(a, A[i], d, x, y);
         if (dr % d) return -;
         //cout << x << ' ' << y << ' ' << d << endl;
         LL tmp = a * x;
         a = lcm(a, A[i]);
         tmp = (tmp % a + a) % a * dr / d + r;
         r = (tmp % a + a) % a;
         //cout << a << ' ' << r << ' ' << tmp << endl;
     }
     return r ? r : a;
 }

 int main() {
     int n, T, cas = ;
     cin >> T;
     while (T-- && cin >> n) {
         for (int i = ; i < n; i++) cin >> A[i];
         for (int i = ; i < n; i++) cin >> R[i];
         cout << "Case " << cas++ << ": " << work(n) << endl;
     }
     return ;
 }

Problem - 1573

 #include <cstdio>
 #include <iostream>
 #include <cmath>
 #include <algorithm>
 #include <cstring>

 using namespace std;

 typedef long long LL;
 inline LL gcd(LL a, LL b) { return b ? gcd(b, a % b) : a;}
 inline LL lcm(LL a, LL b) { return a / gcd(a, b) * b;}
 void gcd(LL a, LL b, LL &d, LL &x, LL &y) {
     if (b) { gcd(b, a % b, d, y, x); y -= a / b * x;}
     else d = a, x = , y = ;
 }

 LL A[], R[], LCM;
 LL work(int n) {
     LL d, x, y, a = A[], r = R[];
     for (int i = ; i < n; i++) {
         LL dr = R[i] - r;
         gcd(a, A[i], d, x, y);
         //cout << d << ' ' << x << ' ' << y << endl;
         if (dr % d) return -;
         LL tmp = a * x * dr / d + r;
         a = lcm(a, A[i]);
         r = (tmp % a + a) % a;
         //cout << a << '~' << r << endl;
     }
     LCM = a;
     return r ? r : a;
 }

 int main() {
     //freopen("in", "r", stdin);
     int T, n;
     LL r;
     cin >> T;
     while (cin >> r >> n) {
         for (int i = ; i < n; i++) cin >> A[i];
         for (int i = ; i < n; i++) cin >> R[i];
         LL ans = work(n);
         //cout << ans << ' ' << LCM << endl;
         if (~ans) cout << max(0ll, (r - ans + LCM) / LCM) << endl;
         else puts("");
     }
 }

Problem - 1930

 #include <cstdio>
 #include <iostream>
 #include <cmath>
 #include <algorithm>
 #include <cstring>

 using namespace std;

 typedef long long LL;
 inline LL gcd(LL a, LL b) { return b ? gcd(b, a % b) : a;}
 inline LL lcm(LL a, LL b) { return a / gcd(a, b) * b;}
 void gcd(LL a, LL b, LL &d, LL &x, LL &y) {
     if (b) { gcd(b, a % b, d, y, x); y -= a / b * x;}
     else d = a, x = , y = ;
 }

 LL A[], R[];

 bool check(LL a) {
     for (int i = ; i < ; i++) {
         if (a %  >  || a %  <= ) return false;
         a /= ;
     }
     return true;
 }

 LL cal() {
     LL d, x, y, a = A[], r = R[];
     for (int i = ; i < ; i++) {
         LL dr = R[i] - r;
         gcd(a, A[i], d, x, y);
         if (dr % d) {
             puts("WTF??");
             while () ;
         }
         LL tmp = a * x;
         //a = lcm(a, A[i]);
         a *= A[i];
         tmp %= a;
         tmp *= dr / d;
         tmp += r;
         r = (tmp % a + a) % a;
         //cout << a << '~' << r << endl;
     }
     while (!check(r)) r += a;
     //cout << r << ' ' << a << endl;
     return r;
 }

 LL cal(LL x) {
     for (int i = ; i < ; i++) R[i] = x % , x /= ;
     //for (int i = 0; i < 4; i++) cout << A[i] << ' ' << R[i] << endl;
     return cal();
 }

 const LL ep[] = { , , };
 char out[];

 int main() {
     //freopen("in", "r", stdin);
     int T, n, p;
     LL tmp, x;
     cin >> T;
     while (T-- && cin >> n) {
         p = ;
         for (int i = ; i < ; i++) cin >> A[i];
         reverse(A, A + );
         for (int i = ; i < n; i++) {
             cin >> tmp;
             tmp = cal(tmp);
             for (int i = ; i < ; i++) {
                 x = tmp / ep[i] % ;
                 out[p++] = x ==  ? ' ' : (x -  + 'A');
             }
         }
         out[p] = ;
         while (p >  && out[p - ] == ' ') out[--p] = ;
         puts(out);
     }
     return ;
 }

——written by Lyon