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7.11

时间:2023-07-11 20:45:33浏览次数:35  
标签:7.11 构造方法 int num include num1

#include<iostream>
#include<math.h>
using namespace std;
int num1(int a) {
    int t = a;
    int num = 1;//纪录输入数据的位数
    for (int i = 0; i < 100; i++) {
        if (t / 10 > 0) {
            num++;
            t /= 10;
        }
        else {
            break;
        }
    }
        return num;
    }
void show(int a){
    int t = a;
    int n;
    int num = num1(a);
    while (num-- > 0) {
        n = a / (int)pow(10.0, num);
        switch (n) {
        case 0:
            cout << " ling";
            break;
        case 1:
            cout << " yi";
            break;
        case 2:
            cout << " er";
            break;
        case 3:
            cout << " san";
            break;
        case 4:
            cout << " si";
            break;
        case 5:
            cout << " wu";
            break;
        case 6:
            cout << " liu";
            break;
        case 7:
            cout << " qi";
            break;
        case 8:
            cout << " ba";
            break;
        case 9:
            cout << " jiu";
            break;
        }
        a = a % (int)pow(10.0, num);
    }
    cout << "\n";
}

int main() {
    int a;//输入的数据
    int t;//暂时储存a的值
    cin >> a;
    t = a;
    if (a < 0) {
        cout << "fu";
        a *= (-1);
        show(a);
    }
    else {
        int num = num1(a);
        int n = a / (int)pow(10.0, (num-1));
        switch (n) {
        case 0:
            cout << "ling";
            break;
        case 1:
            cout << "yi";
            break;
        case 2:
            cout << "er";
            break;
        case 3:
            cout << "san";
            break;
        case 4:
            cout << "si";
            break;
        case 5:
            cout << "wu";
            break;
        case 6:
            cout << "liu";
            break;
        case 7:
            cout << "qi";
            break;
        case 8:
            cout << "ba";
            break;
        case 9:
            cout << "jiu";
            break;
        }
        a = a % (int)pow(10.0,(num-1));

        show(a);
    }

    return 0;
}

 这个题目一直没能完全实现,难点在于他的输出格式和适用性

我打的这个代码虽然格式正确但是适用性不足,如果出现1025这种数据中间有0存在就无法输出正确结果

网课今天看了类的构造方法和C不同的是JAVA在构造方法前也要加上public

还看了私有成员,私有成员的每一个成员前都要加上private

标签:7.11,构造方法,int,num,include,num1
From: https://www.cnblogs.com/xuxingkai/p/17545866.html

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