不定方程与扩展欧几里得
时间不多了，先把代码放上来。
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 5 using namespace std; 6 7 typedef long long ll; 8 9 ll gcd(ll a, ll b) 10 { 11 return (b == 0)?a:(gcd(b, a%b)); 12 } 13 14 ll exgcd(ll a, ll b, ll &x, ll &y) 15 { 16 if(b == 0){ 17 x = 1; y = 0; 18 return a; 19 } 20 ll t = exgcd(b, a%b, y, x); 21 y -= (a/b)*x; 22 23 return t; 24 } 25 26 int main() 27 { 28 ll x, y; 29 ll a, b, m; 30 cin>>a>>b>>m; 31 // 有解的充要条件 32 if(m%gcd(a, b) == 0) cout<<"Yes\n"; 33 ll s = exgcd(a, b, x, y); 34 x = x*m/s; 35 y = y*m/s; 36 for(int t = -100;t<=300;t++){ 37 printf("%lld*(%lld) + %lld*(%lld) = %lld\n", a, x, b, y, m); 38 /* 39 x -= b/s; 40 y += a/s; 41 */ 42 // 通解, 可往左右走 43 x -= b/s; 44 y += a/s; 45 } 46 47 return 0; 48 }
