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 = 11;
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 = 1, y = 0;}
}

LL work(int n) {
    LL x, y, d, a = A[0], r = R[0];
    for (int i = 1; i < n; i++) {
        LL dr = R[i] - r;
        gcd(a, A[i], d, x, y);
        if (dr % d) return -1;
        //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 = 1;
    cin >> T;
    while (T-- && cin >> n) {
        for (int i = 0; i < n; i++) cin >> A[i];
        for (int i = 0; i < n; i++) cin >> R[i];
        cout << "Case " << cas++ << ": " << work(n) << endl;
    }
    return 0;
}

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 = 1, y = 0;
}

LL A[11], R[11], LCM;
LL work(int n) {
    LL d, x, y, a = A[0], r = R[0];
    for (int i = 1; i < n; i++) {
        LL dr = R[i] - r;
        gcd(a, A[i], d, x, y);
        //cout << d << ' ' << x << ' ' << y << endl;
        if (dr % d) return -1;
        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 = 0; i < n; i++) cin >> A[i];
        for (int i = 0; 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("0");
    }
}

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 = 1, y = 0;
}

LL A[6], R[6];

bool check(LL a) {
    for (int i = 0; i < 3; i++) {
        if (a % 100 > 27 || a % 100 <= 0) return false;
        a /= 100;
    }
    return true;
}

LL cal() {
    LL d, x, y, a = A[0], r = R[0];
    for (int i = 1; i < 4; i++) {
        LL dr = R[i] - r;
        gcd(a, A[i], d, x, y);
        if (dr % d) {
            puts("WTF??");
            while (1) ;
        }
        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 = 0; i < 4; i++) R[i] = x % 100, x /= 100;
    //for (int i = 0; i < 4; i++) cout << A[i] << ' ' << R[i] << endl;
    return cal();
}

const LL ep[3] = { 10000, 100, 1};
char out[111111];

int main() {
    //freopen("in", "r", stdin);
    int T, n, p;
    LL tmp, x;
    cin >> T;
    while (T-- && cin >> n) {
        p = 0;
        for (int i = 0; i < 4; i++) cin >> A[i];
        reverse(A, A + 4);
        for (int i = 0; i < n; i++) {
            cin >> tmp;
            tmp = cal(tmp);
            for (int i = 0; i < 3; i++) {
                x = tmp / ep[i] % 100;
                out[p++] = x == 27 ? ' ' : (x - 1 + 'A');
            }
        }
        out[p] = 0;
        while (p > 0 && out[p - 1] == ' ') out[--p] = 0;
        puts(out);
    }
    return 0;
}

——written by Lyon