欧几里德算法

上次做2017 apac test 最后一轮的时候，第三题需要使用扩展欧几里德算法来求解方程的解的个数，刚好学习一下这个算法。上次学了一下，忘记记录下来，这次在做hr的题目的时候，有需要这个算法，还是不熟练，这里还是把上次的代码，先贴出来吧，需要的进行研究！

  1 #include<bits/stdc++.h>
  2 #define pb push_back
  3 #define FOR(i, n) for (int i = 0; i < (int)n; ++i)
  4 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl
  5 typedef long long ll;
  6 using namespace std;
  7 typedef pair<int, int> pii;
  8 const int maxn = 1e3 + 10;
  9 int a, b;
 10 int gcd(int x, int y) {
 11     if(y == 0) return x;
 12     return gcd(y, x % y);
 13 }
 14 int gcd2(int x, int y) {
 15     while(y != 0) {
 16         int t = x % y;
 17         x = y;
 18         y = t;
 19     }
 20     return x;
 21 }
 22 int exgcd(int a, int b, int &x, int &y) {
 23     if(b == 0) {
 24         x = 1;
 25         y = 0;
 26         return a;
 27     }
 28     int r = exgcd(b, a % b, x, y);
 29     int t = x;
 30     x = y;
 31     y = t - a / b * y;
 32     return r;
 33 }
 34 int work(int a, int b, int c) {
 35     int x, y;
 36     ll t = exgcd(a, b, x, y);
 37     //cout << t << " " << x << " " << y << endl;
 38     if(c % t != 0) return 0;
 39     ll f = c / t;
 40     //double z1 = 1.0 * -x * f * t / b;
 41     //double z2 = 1.0 * y * f * t / a;
 42     //int x1 = z1, x2 = z2;
 43     ll t1 = floor(1.0 * -x * f * t / b);
 44     ll t2 = floor(1.0 * y * f * t / a);
 45     //cerr << z1 << " " << x1 << " " << t1 << " " << z2 << " " << x2 << " " << t2 << endl;
 46     if(1ll * y * f * t % a == 0) t2--;
 47     return t2 - t1 < 0 ? 0 : t2 - t1;
 48 }
 49 /*
 50 cin >> a >> b;
 51     cout << __gcd(a, b) << endl;
 52     cout << gcd(a, b) << " " << gcd2(a, b) << endl;
 53     int x, y;
 54     int t = exgcd(a, b, x, y);
 55     cout << t << " " << x << " " << y << endl;
 56 */
 57 void extgcd(ll a,ll b,ll& d,ll& x,ll& y){
 58     if(!b){
 59         d=a;x=1;y=0;
 60     }else{
 61         extgcd(b,a%b,d,y,x);
 62         y-=x*(a/b);
 63     }
 64 }
 65 
 66 int work1(int a,int b,int n){
 67     if(n%__gcd(a,b)!=0)return 0;
 68     ll g,p0,q0;
 69     extgcd(a,b,g,p0,q0);
 70     p0*=n/g;q0*=n/g;
 71     int less=floor((-p0)*g*1.0/b),great=floor(q0*g*1.0/a);
 72     if(q0*g%a==0)--great;
 73     return great-less<0?0:great-less;
 74 }
 75 void solve1() {
 76     cin >> a >> b;
 77     ll res = 0;
 78     //if(a % b == 0) res++;
 79     for (int i = b; i <= a; i += b) {
 80         int now = a;
 81         if(now % i == 0) res++;
 82         res += work(i, i + 1, now);
 83         res += work(i, i + 2, now);
 84         for (int j = 1; ; j++) {
 85             int t = now - i * j;
 86             if(t <= 0) break;
 87             res += work(i + 1, i + 2, t);
 88         }
 89         //cout << i << " " << res << endl;
 90     }
 91     cout << res << endl;
 92 }
 93 ll work2(int a, int b, int c) {
 94     int x, y;
 95     ll t = exgcd(a, b, x, y);
 96     //cout << t << " " << x << " " << y << endl;
 97     if(c % t != 0) return 0;
 98     ll f = c / t;
 99     //double z1 = 1.0 * -x * f * t / b;
100     //double z2 = 1.0 * y * f * t / a;
101     //int x1 = z1, x2 = z2;
102     ll t1 = floor(1.0 * -x * f * t / b);
103     ll t2 = floor(1.0 * y * f * t / a);
104     //cerr << z1 << " " << x1 << " " << t1 << " " << z2 << " " << x2 << " " << t2 << endl;
105     if(1ll * -x * f * t % b == 0) t1--;
106     return t2 - t1 < 0 ? 0 : t2 - t1;
107 }
108 void solve() {
109     cin >> a >> b;
110     ll res = 0;
111     //if(a % b == 0) res++;
112     for (int i = b; i <= a; i += b) {
113         //int now = a - b;
114         for (int j = 1; ; j++) {
115             int now = a - i * j;
116             if(now < 0) break;
117             res += work2(i + 1, i + 2, now);
118         }
119 
120         //cout << i << " " << res << endl;
121     }
122     cout << res << endl;
123 }
124 int main() {
125     freopen("test.in", "r", stdin);
126     freopen("test.out", "w", stdout);
127     //freopen("test.err", "w", stderr);
128     //cout << __builtin_popcount(11) << endl;
129     //return 1;
130     //cout << work(2, 3, 5) << endl;
131     //return 0;
132     int _;
133     cin >> _;
134     for (int i = 1; i <= _; i++) {
135         cout << "Case #" << i << ": ";
136         solve();
137     }
138     return 0;
139 }

 

转载于:https://www.cnblogs.com/y119777/p/6126389.html