HDU 3579 中国剩余定理模数不互质
这里用到了求解模线性方程组的一种一般方法：合并法。而中国剩余定理要求模数两两互素。
合并方法的说明：转自math.exchange。还是英文说得清楚。。
代码：
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 const int maxn = 1000+10; 5 6 class Extend_Euclid{ 7 8 private: 9 int size_e = 0; // number of equations 10 int max_size; // 11 int a[maxn]; 12 int m[maxn]; 13 14 /* 15 * no need to make sure that a is nl then b 16 */ 17 ll gcd(ll a, ll b){ 18 return b ? gcd(b, a%b) : a; 19 } 20 21 /* 22 * if you want the minimal non-neg x, (x mod b+b) mod b is the answer 23 */ 24 void extend_Euclid(ll a, ll b, ll &x, ll &y){ 25 if(b==0){ 26 x = 1; 27 y = 0; 28 return; 29 } 30 extend_Euclid(b, a%b, x, y); 31 ll tmp = x; 32 x = y; 33 y = tmp-(a/b)*y; 34 } 35 36 bool merge(ll a1, ll m1, ll a2, ll m2, ll &new_a, ll &new_m){ 37 38 if((a1-a2) % gcd(m1, m2) != 0){ 39 //cout <<"here: " <<a1-a2 <<' ' << gcd(m1, m2) <<'\n'; 40 return false; 41 } 42 43 ll x=0, y=0; 44 extend_Euclid(m1, m2, x, y); 45 new_a = a1 - m1/gcd(m1, m2)*x*(a1-a2); 46 new_m = m1*m2/gcd(m1, m2); 47 //new_a = a2 + m2/gcd(m1, m2)*y*(a1-a2); 48 new_a = (new_a%new_m + new_m) % new_m; 49 return true; 50 } 51 52 public: 53 Extend_Euclid(int size){ 54 size_e = size; 55 } 56 57 void read(){ 58 for (int i = 0; i < size_e; ++i){ 59 cin >>m[i]; 60 } 61 for (int i = 0; i < size_e; ++i){ 62 cin >>a[i]; 63 } 64 } 65 66 ll solve(){ 67 //cout <<gcd(14, 57) <<'\n'; 68 //merge equation i and i-1 69 ll new_a=a[0], new_m=m[0]; 70 for (int i = 1; i < size_e; ++i){ 71 if(!merge(new_a, new_m, a[i], m[i], new_a, new_m)){ 72 return -1; 73 } 74 } 75 ll res = (new_a%new_m + new_m) % new_m; 76 if(res==0) return new_m; 77 else return res; 78 } 79 80 }; 81 82 int main(){ 83 int T, n, cnt=0; 84 cin >>T; 85 while(T--){ 86 cin >>n; 87 Extend_Euclid ee(n); 88 ee.read(); 89 ll res = ee.solve(); 90 cout <<"Case " <<++cnt <<": "; 91 92 cout <<res <<'\n'; 93 94 } 95 return 0; 96 }
