// Original C++ solution (slower):
#include <iostream>

#include <sstream>

#include <string>

#include <algorithm>

#include <vector>

#include <stack>

#include <queue>

#include <set>

#include <map>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <cassert>



using namespace std;



#define FOR(i,k,n) for(int i=(k); i<(int)(n); ++i)

#define REP(i,n) FOR(i,0,n)

#define FORIT(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)



template<class T> void debug(T begin, T end){ for(T i = begin; i != end; ++i) cerr<<*i<<" "; cerr<<endl; }

inline bool valid(int x, int y, int W, int H){ return (x >= 0 && y >= 0 && x < W && y < H); }



typedef long long ll;

const int INF = 100000000;

const double EPS = 1e-8;

const int MOD = 1000000007;

int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};

int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

// 2次方程式 ax^2 + bx + c = 0の解 (重解を一つにまとめる)

vector<double> quadratic(double a, double b, double c){

  // assert(a != 0);

  if(abs(a) < EPS){

    if(abs(b) < EPS) return vector<double>();

    return vector<double>(1, -c / b);

  }

  double D = b*b - 4*a*c;

  if(D < 0) return vector<double>();

  if(D == 0) return vector<double>(1, -b/(2.0 * a));



  // |b| >> |ac|の時の桁落ちを避けるために

  // x_1 = (-b-sign(b)*sqrt(D))/(2*a), x_2 = c / (a*x_1)を利用する

  vector<double> res;

  int sign = (b >= 0) ? 1 : -1;

  double x1 = (-b - sign * sqrt(D))/(2.0 * a);

  double x2 = c / (a * x1);

  res.push_back(x1);

  res.push_back(x2);

  return res;

}

double f(double a, double b, double c, double x){

  return a * x * x + b * x + c ;

}

double g(double A, double B, double x){

  return sqrt(1 + (A * x + B) * (A * x + B));

}



int main(){

  double W, H;

  int N;

  while(cin >> W >> H >> N){

    double a[50], b[50], c[50];

    vector<double> inter_point;

    REP(i, N){

      double p, q;

      cin >> a[i] >> p >> q;

      b[i] = -2 * a[i] * p;

      c[i] = a[i] * p * p + q;

      vector<double> v = quadratic(a[i], b[i], c[i]);

      REP(i, v.size()){

        double x = v[i];

        x = max(0.0, x);

        x = min(W, x);

        bool ok = true;

        REP(i, inter_point.size()) if(abs(x - inter_point[i]) < EPS) ok = false;

        if(ok) inter_point.insert(lower_bound(inter_point.begin(), inter_point.end(), x), x);

      }

    }

    REP(i, N) FOR(j, i + 1, N) {

      vector<double> v = quadratic(a[i] - a[j], b[i] - b[j], c[i] - c[j]);

      REP(k, v.size()){

        double x = v[k];

        x = max(0.0, x);

        x = min(W, x);

        bool ok = true;

        REP(i, inter_point.size()) if(abs(x - inter_point[i]) < EPS) ok = false;

        if(ok) inter_point.insert(lower_bound(inter_point.begin(), inter_point.end(), x), x);

      }

    }

    double ans = 0;

    REP(i, inter_point.size() - 1){

      double lb = inter_point[i], ub = inter_point[i + 1];

      double mid = (lb + ub) / 2;

      int idx = 0;

      REP(j, N) if(f(a[idx], b[idx], c[idx], mid) < f(a[j], b[j], c[j], mid)){

        idx = j;

      }

      if(!(f(a[idx], b[idx], c[idx], lb + EPS) >= - EPS)) continue;

      if(!(f(a[idx], b[idx], c[idx], ub - EPS) >= - EPS)) continue;

      //printf("[%lf, %lf] idx = %d\n", lb, ub, idx);

      double A = 2 * a[idx];

      double B = b[idx];

      const int N = 10000;

      double delta = (ub - lb) / N;

      double S = 0;

      S += g(A, B, lb) + g(A, B, ub);

      for(int i = 1; i < N; i += 2){

        S += 4.0 * g(A, B, lb + delta * i);

      }

      for(int i = 2; i < N; i += 2){

        S += 2.0 * g(A, B, lb + delta * i);

      }

      ans += S * delta / 3.0;

    }

    assert(!isnan(ans));

    printf("%.16lf\n", ans);

  }

  return 0;

}

// Optimized solution to the same problem (faster):
#include <iostream>

#include <sstream>

#include <string>

#include <algorithm>

#include <vector>

#include <stack>

#include <queue>

#include <set>

#include <map>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <cassert>



using namespace std;



#define FOR(i,k,n) for(int i=(k); i<(int)(n); ++i)

#define REP(i,n) FOR(i,0,n)

#define FORIT(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)



template<class T> void debug(T begin, T end){ for(T i = begin; i != end; ++i) cerr<<*i<<" "; cerr<<endl; }

inline bool valid(int x, int y, int W, int H){ return (x >= 0 && y >= 0 && x < W && y < H); }



typedef long long ll;

const int INF = 100000000;

const double EPS = 1e-8;

const int MOD = 1000000007;

int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};

int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

// 2次方程式 ax^2 + bx + c = 0の解 (重解を一つにまとめる)

vector<double> quadratic(double a, double b, double c){

  // assert(a != 0);

  if(abs(a) < EPS){

    if(abs(b) < EPS) return vector<double>();

    return vector<double>(1, -c / b);

  }

  double D = b*b - 4*a*c;

  if(D < 0) return vector<double>();

  if(D == 0) return vector<double>(1, -b/(2.0 * a));



  // |b| >> |ac|の時の桁落ちを避けるために

  // x_1 = (-b-sign(b)*sqrt(D))/(2*a), x_2 = c / (a*x_1)を利用する

  vector<double> res;

  int sign = (b >= 0) ? 1 : -1;

  double x1 = (-b - sign * sqrt(D))/(2.0 * a);

  double x2 = c / (a * x1);

  res.push_back(x1);

  res.push_back(x2);

  return res;

}

double f(double a, double b, double c, double x){

  return a * x * x + b * x + c ;

}

double g(double A, double B, double x){

  return sqrt(1 + (A * x + B) * (A * x + B));

}



int main(){

  double W, H;

  int N;

  while(cin >> W >> H >> N){

    double a[50], b[50], c[50];

    vector<double> inter_point;

    REP(i, N){

      double p, q;

      cin >> a[i] >> p >> q;

      b[i] = -2 * a[i] * p;

      c[i] = a[i] * p * p + q;

      vector<double> v = quadratic(a[i], b[i], c[i]);

      REP(i, v.size()){

        double x = v[i];

        x = max(0.0, x);

        x = min(W, x);

        bool ok = true;

        REP(i, inter_point.size()) if(abs(x - inter_point[i]) < EPS) ok = false;

        if(ok) inter_point.insert(lower_bound(inter_point.begin(), inter_point.end(), x), x);

      }

    }

    REP(i, N) FOR(j, i + 1, N) {

      vector<double> v = quadratic(a[i] - a[j], b[i] - b[j], c[i] - c[j]);

      REP(k, v.size()){

        double x = v[k];

        x = max(0.0, x);

        x = min(W, x);

        bool ok = true;

        REP(i, inter_point.size()) if(abs(x - inter_point[i]) < EPS) ok = false;

        if(ok) inter_point.insert(lower_bound(inter_point.begin(), inter_point.end(), x), x);

      }

    }

    double ans = 0;

    REP(i, inter_point.size() - 1){

      double lb = inter_point[i], ub = inter_point[i + 1];

      double mid = (lb + ub) / 2;

      int idx = 0;

      REP(j, N) if(f(a[idx], b[idx], c[idx], mid) < f(a[j], b[j], c[j], mid)){

        idx = j;

      }

      if(!(f(a[idx], b[idx], c[idx], lb + EPS) >= - EPS)) continue;

      if(!(f(a[idx], b[idx], c[idx], ub - EPS) >= - EPS)) continue;

      //printf("[%lf, %lf] idx = %d\n", lb, ub, idx);

      double A = 2 * a[idx];

      double B = b[idx];

      const int N = 1000;

      double delta = (ub - lb) / N;

      double S = 0;

      S += g(A, B, lb) + g(A, B, ub);

      for(int i = 1; i < N; i += 2){

        S += 4.0 * g(A, B, lb + delta * i);

      }

      for(int i = 2; i < N; i += 2){

        S += 2.0 * g(A, B, lb + delta * i);

      }

      ans += S * delta / 3.0;

    }

    assert(!isnan(ans));

    printf("%.16lf\n", ans);

  }

  return 0;

}

// Fastest known solution for this problem: