编程打卡：C语言趣味编程习题做

数制转换

问题描述

给定一个M进制的数x，实现对x向任意非M进制的数的转换。

设计思路

输入M进制的数x，将x转换为十进制数，再将十进制数转换为任意非M进制的数。

流程图

graph A["开始"] --> B["输入M进制的数x"] -->C["将x转换为十进制数"] --> D["将十进制数转换为任意非M进制的数"] --> E["输出"] --> F["结束"]

代码实现

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int num2dec(int m,string num)   {
    int tmpm = 1;
    int sum = 0;
    for (int i = num.length()-1; i >= 0; i--) {
        if (num[i] >= '0' && num[i] <= '9') {
            sum += (num[i] - '0') * tmpm;
            tmpm *= m;
        }
        else if (num[i] >= 'A' && num[i] <= 'F') {
            sum += (num[i] - 'A' + 10) * tmpm;
            tmpm *= m;
        }
        else if (num[i] >= 'a' && num[i] <= 'f') {
            sum += (num[i] - 'a' + 10) * tmpm;
            tmpm *= m;
        }
    }
    return sum;
}
string dec2num(int m,int num) {
    string res;
    while (num) {
        int tmp = num % m;
        if (tmp >= 0 && tmp <= 9)   {
            res += tmp + '0';
        }
        else {
            res += tmp - 10 + 'A';
        }
        num /= m;
    }
    reverse(res.begin(),res.end());
    return res;
}
int main()  {
    int m,t;
    cin >> m >> t;
    string num;
    cin >> num;
    cout << dec2num(t,num2dec(m,num));
}

运行结果

输入

8 16
1234

输出

29C

输入

16 10
ff

输出

个人所得税问题

问题描述

编写程序，计算个人所得税。

设计思路

用数组存储不同工资阶段区间以及个人所得税率，根据输入的金额，循环判断来计算个人所得税

流程图

graph A[开始] --> B[输入工资] --> C{工资>3500}; C -- No --> D[输出个人所得税为0] --> E[结束]; C -- Yes --> F[i = 0] --> G{"工资>个税[i].min?"} -- Yes --> H{"工资<=个税[i].max?"} -- Yes --> I[计算输出个人所得税] --> E; H -- No --> J[i++] --> G;

代码实现

#include <iostream>
#include <climits>
using namespace std;
struct zone {
    int min;
    int max;
    double rate;
};
struct zone tax[] = {{0,1500,0.03},{1500,4500,0.10},{4500,9000,0.20},{9000,35000,0.25},{35000,55000,0.30},{55000,80000,0.35},{80000,INT_MAX,0.45}};
int main () {
    double payTax = 0,salary;
    cin >> salary;
    if (salary <= 3500)
        cout << 0 << endl;
    else {
        salary -= 3500;
        for (struct zone i : tax)   {
            cout << salary << " " << i.min << " " << i.max << " " << i.rate << endl;
            if (salary > i.max) {
                payTax += (i.max - i.min) * i.rate;
            }
            else {
                payTax += (salary - i.min) * i.rate;
                break;
            }
        }
        cout << payTax << endl;
    }
}