P1883 函数

R43928510

problem/P1883/P1883.cpp

@@ -0,0 +1,40 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int t, n, a[10005], b[10005], c[10005];
+
+double f(double x, int i) {
+    return x * x * a[i] + x * b[i] + c[i];
+}
+
+double check(double x) {
+    double ans = f(x, 1);
+    for (int i = 2; i <= n; i++) {
+        ans = max(ans, f(x, i));
+    }
+    return ans;
+}
+
+int main() {
+    cin >> t;
+    while (t--) {
+        cin >> n;
+        for (int i = 1; i <= n; i++) {
+            cin >> a[i] >> b[i] >> c[i];
+        }
+        double l = 0.00, r = 1000.00, mid1, mid2;
+        while (r - l > 1e-11) {
+            mid1 = l + (r - l) / 3.0;
+            mid2 = r - (r - l) / 3.0;
+            if (check(mid1) > check(mid2)) {
+                l = mid1;
+            }
+            else {
+                r = mid2;
+            }
+        }
+        cout << fixed << setprecision(4) << check(l) << endl;
+    }
+    return 0;
+}