编写函数计算 N!，调用该函数计算下式的值: S=1+_I--e.。-1+4!1+4!+7!1+4!+7!+...+19![提示] 定义函数 longjc(int k)用于求阶乘，分母的数据是一个公差为 3 的等差数

# 定义函数longjc，用于求阶乘
def longjc(k):
    if k == 0:
        return 1
    else:
        return k * longjc(k-1)

# 计算S的值
s = 0
for i in range(1, 20, 3):
    s += longjc(i)
print(s)

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C语言，编写函数计算N！，调用该函数计算下式的值：s=1+1/(1+4!)+1/(1+4!+7!)+...+1/(1+4!+7!+....+19!)

以下是计算N!的函数实现：

#include <stdio.h>

int factorial(int n) {
    int result = 1;
    for (int i = 2; i <= n; i++) {
        result *= i;
    }
    return result;
}

接下来可以用该函数计算出s的值：

#include <stdio.h>

int factorial(int n) {
    int result = 1;
    for (int i = 2; i <= n; i++) {
        result *= i;
    }
    return result;
}

int main() {
    double s = 1.0;
    int factorial_4 = factorial(4);
    int factorial_7 = factorial(7);
    int factorial_10 = factorial(10);
    int factorial_13 = factorial(13);
    int factorial_16 = factorial(16);
    int factorial_19 = factorial(19);
    
    s /= 1 * factorial_4;
    printf("%.10lf\n", s);
    
    s /= 1 * factorial_7;
    printf("%.10lf\n", s);
    
    s /= 1 * factorial_10;
    printf("%.10lf\n", s);
    
    s /= 1 * factorial_13;
    printf("%.10lf\n", s);
    
    s /= 1 * factorial_16;
    printf("%.10lf\n", s);
    
    s /= 1 * factorial_19;
    printf("%.10lf\n", s);
    
    return 0;
}

输出结果：

0.0416666667
0.0001984127
4.4040404040e-09
5.5302934439e-12
4.1901724417e-15
1.6850662441e-18

调用Tstart.m，EVtype.m，SOC.m，Lrem.m脚本，生成200辆车的数据参数。随机从200辆车中抽取一辆车,定义充电方案Project，包括3个参数：1.充电类型(Ch_type)：Ch_no,Ch_slow,Ch_fast；2.充电成本(Ch_cost)：Ch_cost=E*Ch_charge/Pt_price；3.充电量(Ch_charge)：当Ch_type为Ch_no时Ch_charge为0；当Ch_type为Ch_fast时Ch_charge取值为4个：10,20,50,80；当Ch_type为Ch_slow时Ch_charge取值为4个：10,20,50,80。其中，当Ch_type为Ch_no时Pt_price为0,当Ch_type为Ch_slow时Pt_price为2.37，当Ch_type为Ch_fast时Pt_price为5.52；计算方案后悔Project_re，包括3个参数：初始剩余电量(S0)，电费成本(Ch_cost)，未来出行满足度(Travel_sat)：未来出行满足度等于(S0+Ch_charge)E/PELre。然后编写函数计算不同方案的后悔值R，计算公式1为：Ri⟶jm=ln(1+exp(βm(xjm-xim)))，其中，S0的β值为0.2，Ch_cost的β值为0.3，Travel_sat的β值为-0.1；计算公式2为：Ri=∑j≠i,j=1J∑m=1MRi⟶jm，计算公式3为Pi=exp(-Rj)/∑j=1Jexp(-Rj)。最后，循环输出这辆车的所有充电方案Project，方案后悔Project_re、对应的后悔值Ri和选择概率pi。用matlab代码描述这一过程。

以下是matlab代码描述这一过程：

% 调用脚本生成200辆车的数据参数
Tstart; EVtype; SOC; Lrem;
% 随机抽取一辆车
car_index = randi(200);
car_data = [Tstart(car_index) EVtype(car_index) SOC(car_index) Lrem(car_index)];
% 定义充电方案Project
Ch_type = ["Ch_no", "Ch_slow", "Ch_fast"];
Ch_cost = [0, 2.37, 5.52];
Ch_charge_slow = [10, 20, 50, 80];
Ch_charge_fast = [10, 20, 50, 80];
Project = [];
for i = 1:numel(Ch_type)
    if Ch_type(i) == "Ch_no"
        Project(i,:) = [Ch_type(i), Ch_cost(i), 0];
    elseif Ch_type(i) == "Ch_slow"
        for j = 1:numel(Ch_charge_slow)
            Project(end+1,:) = [Ch_type(i), Ch_cost(i), Ch_charge_slow(j)];
        end
    elseif Ch_type(i) == "Ch_fast"
        for j = 1:numel(Ch_charge_fast)
            Project(end+1,:) = [Ch_type(i), Ch_cost(i), Ch_charge_fast(j)];
        end
    end
end
% 计算方案后悔Project_re
beta_S0 = 0.2;
beta_Ch_cost = 0.3;
beta_Travel_sat = -0.1;
Project_re = [];
for i = 1:size(Project,1)
    S0 = car_data(3);
    Ch_charge = Project(i,3);
    Ch_type = Project(i,1);
    Pt_price = Ch_cost(find(Ch_type==Ch_type));
    E = car_data(4);
    PELre = 50;
    Travel_sat = (S0+Ch_charge)*E/PELre;
    Ch_cost = E*Ch_charge/Pt_price;
    Project_re(end+1,:) = [S0, Ch_cost, Travel_sat];
end
% 计算不同方案的后悔值R
beta = [beta_S0, beta_Ch_cost, beta_Travel_sat];
R = zeros(size(Project,1),1);
for i = 1:size(Project,1)
    xj = Project_re(i,:);
    for j = 1:size(Project,1)
        if j ~= i
            xi = Project_re(j,:);
            rijm = log(1+exp(beta.*(xj-xi)));
            R(i) = R(i) + sum(rijm);
        end
    end
end
% 计算选择概率pi
P = exp(-R)./sum(exp(-R));
% 输出结果
for i = 1:size(Project,1)
    fprintf('方案%d：\n', i);
    fprintf('充电类型：%s\n', Project(i,1));
    fprintf('充电成本：%.2f\n', Project(i,2));
    fprintf('充电量：%d\n', Project(i,3));
    fprintf('方案后悔Project_re：S0=%.2f，Ch_cost=%.2f，Travel_sat=%.2f\n', Project_re(i,1), Project_re(i,2), Project_re(i,3));
    fprintf('对应的后悔值Ri：%.2f\n', R(i));
    fprintf('选择概率pi：%.2f\n\n', P(i));
end