SGU119 Magic Pairs

题目大意

找出所有的A和B，满足：
∀X,Y∈ℤ，若A0X+B0Y≡0 (mod N)，则AX+BY≡0 (mod N)

算法思路

AX+BY≡0 (mod N)⟺A0B≡B0A (mod N)

∵AX+BY≡0 (mod N)
∴A0(AX+BY)≡0 (mod N)，即A0AX+A0BY≡0 (mod N)
由A0X+B0Y≡0 (mod N)可得，A0X≡−B0Y (mod N)
带入上式得，−B0AY+A0BY≡0 (mod N)
∵Y的取值不全为0，∴A0B≡B0A (mod N)

至此，可以枚举A的值，通过逆元计算对应的B

时间复杂度: O(NlogN)

代码

/** * Copyright © 2015 Authors. All rights reserved. *  * FileName: 119.cpp * Author: Beiyu Li <[e-mail]> * Date: 2015-05-30 */#include <bits/stdc++.h>using namespace std;#define rep(i,n) for (int i = 0; i < (n); ++i)#define For(i,s,t) for (int i = (s); i <= (t); ++i)#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)typedef long long LL;typedef pair<int, int> Pii;const int inf = 0x3f3f3f3f;const LL infLL = 0x3f3f3f3f3f3f3f3fLL;void gcd(int a, int b, int &g, int &x0, int &y0){        if (!b) g = a, x0 = 1, y0 = 0;        else gcd(b, a % b, g, y0, x0), y0 -= x0 * (a / b);}int inv(int a, int m){        int g, x0, y0;        gcd(a, m, g, x0, y0);        return g == 1? (x0 + m) % m: -1;}vector<Pii> vec;void solve(int a0, int b0, int n){        rep(a,n) {                int x = b0 * a % n;                if (a0 % n == 0) {                        if (x) continue;                        rep(b,n) vec.push_back(Pii(a, b));                } else {                        int g = __gcd(a0, n);                        if (x % g != 0) continue;                        int b = inv(a0 / g, n / g) * (x / g) % (n / g);                        while (b < n) vec.push_back(Pii(a, b)), b += n / g;                }        }}int main(){        int n, a0, b0;        scanf("%d%d%d", &n, &a0, &b0);        int g = __gcd(__gcd(a0, b0), n);        solve(a0 / g, b0 / g, n / g);        printf("%d\n", (int)vec.size());        foreach(it,vec) printf("%d %d\n", it->first * g, it->second * g);        return 0;}

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