P5656 【模板】二元一次不定方程 (exgcd)

https://www.luogu.com.cn/record/88197267

Luogu/P5656/P5656.cpp

@@ -0,0 +1,67 @@
+#include <iostream>
+#include <cmath>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+long long exgcd(long long a, long long b, long long& x, long long& y) {
+    if (!b) {
+        x = 1, y = 0;
+        return a;
+    }
+
+    long long g = exgcd(b, a % b, y, x);
+    y -= a / b * x;
+    return g;
+}
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+
+    cin >> t;
+
+    while (t--) {
+        long long a, b, c, x, y;
+
+        cin >> a >> b >> c;
+
+        long long g = exgcd(a, b, x, y);
+
+        if (c % g) {
+            cout << -1 << endl;
+        } else {  // c % g == 0
+            x *= c / g;
+            y *= c / g;
+
+            long long p = b / g,
+                      q = a / g;
+
+            if (x < 0) {
+                long long k = std::ceil((1.0 - x) / p);
+                x += p * k;
+                y -= q * k;
+            } else {  // x >= 0
+                long long k = (x - 1) / p;
+                x -= p * k;
+                y += q * k;
+            }
+
+            if (y > 0) {
+                cout << (y - 1) / q + 1 << ' '
+                     << x << ' '
+                     << (y - 1) % q + 1 << ' '
+                     << x + (y - 1) / q * p << ' '
+                     << y << endl;
+            } else {  // y <= 0
+                cout << x << ' '
+                     << y + q * static_cast<long long>(std::ceil((1.0 - y) / q)) << endl;
+            }
+        }
+    }
+
+    return 0;
+}