itoa函数

char *itoa( int value, char *string,int radix);^[1]

原型说明：

value：欲转换的数据。

string：目标字符串的地址。

radix：转换后的进制数，可以是10进制、16进制等。

程序实例:

#include <stdlib.h>
#include <stdio.h>
int main(void)
{
    int number = 12345;
    char string[32];
    itoa(number, string, 10);
    printf("integer = %d string = %s\n", number, string);
    return 0;
}
/*
设计思路：
1.判断是否有乘除法
2.规定数值范围：通过random乘数值
3.加减有无负数：两个数值随机产生符号
4.除法有无余数：
*/

#include<iostream>
#include<time.h>
#include<string>
# include <stdio.h>
#include<stdlib.h>
#include<iomanip>
#include<cmath>  
using namespace std;

//1.乘除法
int chengchu()
{
   char X;
   cout << "请选择要产生几以内的四则运算：（10 or 100）";
   int range;
   cin >> range;

   cout << "1.是否需要乘除法运算？（Y or N）\n";
   cin >> X;

   if (X == 'Y')
   {

       for (int count = 0; count < 5; count++)
       {
           int a = 0, b = 0;

           a = rand() % range;
           b = rand() % range ;

           //随机产生四则运算符
           int sign = 0;
           sign = (rand() % 100) % 2;
           string opera_sign[2] = { "*", "/" };

           //当b=0且运算为除法时重新生成
           while ((b == 0) & (sign == 1))
           {
               b = rand() % range;
           }

           switch (sign)
           {
           case 0:cout << a << opera_sign[sign] << b << "=" << endl; break;
           case 1:cout << a << opera_sign[sign] << b << "=" << endl; break;
           }
       }
   }
   if (X == 'N')
   {

       for (int count = 0; count < 5; count++)
       {
           int a = 0, b = 0;

           a = rand() % range;
           b = rand() % range;

           //随机产生四则运算符
           int sign = 0;
           sign = (rand() % 100) % 2;
           string opera_sign[2] = { "+", "-" };

           switch (sign)
           {
           case 0:cout << a << opera_sign[sign] << b << "=" << endl; break;
           case 1:cout << a << opera_sign[sign] << b << "=" << endl; break;
           }
       }
   }
   return 0;
}

//2.包含真分数
int fenshu()
{
   char Z;
   cout << "2.是否包含真分数？（Y or N）\n";
   cin >> Z;
   if (Z == 'Y')
   {
       for (int count = 0; count < 5; count++)
       {
           int a1 = 0, b1 = 0, a2 = 0, b2 = 0;

           a1 = rand() % 100;
           b1 = rand() % 100;
           a2 = rand() % 100;
           b2 = rand() % 100;

           //判断是否为真分数
           while (a1>b1 || b1 == 0)
           {
               a1 = rand() % 100;
               b1 = rand() % 100;
           }

           while (a2 > b2 || b2 == 0)
           {
               a2 = rand() % 100;
               b2 = rand() % 100;
           }

           //随机产生四则运算符
           int sign = 0;
           sign = (rand() % 100) % 4;
           string opera_sign[4] = { "+", "-", "*", "/" };

           switch (sign)
           {
           case 0:cout << "(" << a1 << "/" << b1 << ")" << opera_sign[sign] << "(" << a2 << "/" << b2 << ")" << "=" << endl; break;
           case 1:cout << "(" << a1 << "/" << b1 << ")" << opera_sign[sign] << "(" << a2 << "/" << b2 << ")" << "=" << endl; break;
           case 2:cout << "(" << a1 << "/" << b1 << ")" << opera_sign[sign] << "(" << a2 << "/" << b2 << ")" << "=" << endl; break;
           case 3:cout << "(" << a1 << "/" << b1 << ")" << opera_sign[sign] << "(" << a2 << "/" << b2 << ")" << "=" << endl; break;
           }
       }
   }
   if (Z == 'N')
   {
       for (int count = 0; count < 5; count++)
       {
           int a1 = 0, b1 = 0, a2 = 0, b2 = 0;

           a1 = rand() % 100;
           b1 = rand() % 100;
           a2 = rand() % 100;
           b2 = rand() % 100;

           //分母不可为0
           while (b1 == 0)
           {
               b1 = rand() % 100;
           }
           while (b2 == 0)
           {
               b2 = rand() % 100;
           }

           //随机产生四则运算符
           int sign = 0;
           sign = (rand() % 100) % 4;
           string opera_sign[4] = { "+", "-", "*", "/" };

           switch (sign)
           {
           case 0:cout << "(" << a1 << "/" << b1 << ")" << opera_sign[sign] << "(" << a2 << "/" << b2 << ")" << "=" << endl; break;
           case 1:cout << "(" << a1 << "/" << b1 << ")" << opera_sign[sign] << "(" << a2 << "/" << b2 << ")" << "=" << endl; break;
           case 2:cout << "(" << a1 << "/" << b1 << ")" << opera_sign[sign] << "(" << a2 << "/" << b2 << ")" << "=" << endl; break;
           case 3:cout << "(" << a1 << "/" << b1 << ")" << opera_sign[sign] << "(" << a2 << "/" << b2 << ")" << "=" << endl; break;
           }
       }
   }
   return 0;
}

//把数字转换成字符串型
string int_string(int number)
{
   char str1[200];
   itoa(number, str1, 10);
   string str_ = str1;
   return str_;
}

int kuohao()
{
   char K;
   cout << "3.是否包含括号？（Y or N）\n";
   cin >> K;
   if (K == 'Y')
   {
       string stri[4];
       for (int i = 0; i < 4; i++)
       {
           int a = 0, b = 0;
           a = rand() % 100;
           b = rand() % 100;

           int sign = 0;
           sign = (rand() % 100) % 4;
           string opera_sign[4] = { "+", "-", "*", "/" };

           //当b=0且运算为除法时重新生成
           while ((b == 0) & (sign == 3))
           {
               b = rand() % 100;
           }
           string str,ch1, ch2;

           ch1 = int_string(a);
           ch2 = int_string(b);
           //string stri[4];

           switch (sign)
           {
           case 0:
           {
                     str = "(" + ch1 + opera_sign[sign] + ch2 + ")";
           } break;

           case 1:
           {
                     str = "(" + ch1 + opera_sign[sign] + ch2 + ")";
           }break;

           case 2:
           {
                     str = "(" + ch1 + opera_sign[sign] + ch2 + ")";
           } break;

           case 3:
           {
                     str = "(" + ch1 + opera_sign[sign] + ch2 + ")";
           }break;
           }
           stri[i]= str;
       }

       string str5;
   //    str5 = "((" + stri[0] + stri[1] + ")" +stri[2] + ")" + stri[3];
       cout << "((" <<stri[0] <<"+"<<stri[1] << ")*"<<stri[2] <<")" <<"/"<< stri[3];
       }
       
   else
   {
       cout << endl;
   }
      

}
//主函数
int main()
{
   //以现在的系统时间作为随机数的种子来产生随机数
   srand(time(NULL));
   cout << "请您选择四则运算题的条件：\n";
   int result1 = chengchu();
   int result2 = fenshu();
   int result3 = kuohao();
   return 0;
} 