#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <map>
#include <set>
#include <bitset>
#include <sstream>
#include <cstdlib>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <climits>
#include <ctype.h>
using namespace std;

#define PI acos(-1.0)
#define EPS 1e-8
#define MOD 1000000007
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define fi first
#define se second

typedef long double ld;
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<vector<int>> vvi;

struct Matrix {
    vvi data;
    int r, c;
    Matrix(int row, int col, bool identity = false) : r(row), c(col) {
        data.assign(row, vector<int>(col, 0));
        if (identity) {
            for (int i = 0; i < r; ++i) {
                data[i][i] = 1;
            }
        }
    }
    Matrix operator * (Matrix& other) {
        int m = r, n = c, p = other.c;
        Matrix res(m, p);
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < p; ++j) {
                for (int k = 0; k < n; ++k) {
                    res.data[i][j] += data[i][k] * other.data[k][j];
                }
            }
        }
        return res;
    }
};

// 扩展欧几里得算法 res[1] * a + res[2] * b = res[0] = gcd(a, b)
vector<int> extendGcd(int a, int b) {
    if (b == 0) {
        return {a, 1, 0};
    } else {
        vector<int> tmp = extendGcd(b, a % b);
        return {tmp[0], tmp[2], tmp[1] - (a / b) * tmp[2]};
    }
}

// 矩阵快速幂
Matrix matrix_power(Matrix base, ll exp) {
    int n = base.r;
    Matrix res(n, n, true);
    while (exp) {
        if (exp & 1) {
            res = res * base;
        }
        base = base * base;
        exp >>= 1;
    }
    return res;
}

// 带模快速幂
template <typename T>
T power_mod(T base, int exp, int mod = MOD) {
    T res = 1;
    while (exp) {
        if (exp & 1) res = res * base % mod;
        base = base * base % mod;
        exp >>= 1;
    }
    return res;
}

// 快速幂
template <typename T>
T power(T base, int exp) {
    T res = 1;
    while (exp) {
        if (exp & 1) res *= base;
        base *= base;
        exp >>= 1;
    }
    return res;
}

// 求逆元
ll inv(ll a) {
    return power_mod(a, MOD - 2);
}

/******************************** template ********************************/
#define MAXN 105
int n;
int t[MAXN], v[MAXN];

double solve(int s, int e, double upper, double tot) {

    if (s > e) {
        return solve(e, s, upper, tot);
    }
    double t1 = upper - s, t2 = upper - e;
    if (t1 + t2 > tot) {
        upper = 0.5 * (tot + s + e);
    }

    double len1 = upper - s, len2 = upper - e;
    double mid = max(0.0, tot - len1 - len2);
    return 0.5 * (s + upper) * len1 + upper * mid + 0.5 * (e + upper) * len2;
}

int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> t[i];
    }
    for (int i = 1; i <= n; ++i) {
        cin >> v[i];
    }
    double res = 0.0;

    vector<int> dp(n + 1);
    dp[0] = 0;
    dp[n] = 0;

    for (int i = 1; i <= n - 1; ++i) {
        int left = min(dp[i - 1] + t[i], v[i]);
        dp[i] = min(left, v[i + 1]);
    }

    for (int i = n; i >= 2; --i) {
        int right = min(dp[i] + t[i], v[i]);
        dp[i - 1] = min(dp[i - 1], right);
    }

    for (int i = 1; i <= n; ++i) {
        res += solve(dp[i - 1], dp[i], v[i], t[i]);
    }
    printf("%.4lf", res);

    return 0;
}

Measured memory_bytes: 256