题目：
0x02 设计一个复数类型，输入实部和虚部生成一个复数，可进行两个复数求和、两个复数求差、两个复数求积运算。
0x03 用一个整型数组表示10进制大整数，数组的每个元素存储大整数的一位数字，将这个大整数转换为2进制数输出。
代码1：
1 #include <stdio.h> 2 #include<string.h> 3 int Bin2Digit(const char *sSrc, int nSrcLen){ //二进制字符串转十进制数字 4 int num = 0; 5 for(int i=0;i < nSrcLen;i++) //从左到右，i是从左开始的索引，nSrcLen是二进制字符串的长度 6 { 7 num *= 2; //乘2相当于左移一位 8 if(sSrc[i]-'1' == 0) 9 num += 1; 10 else if(sSrc[i]-'0' == 0) 11 num += 0; 12 else{ // 非0非1，不符合要求 13 printf("字符串存在非法字符！"); 14 return -1; 15 } 16 } 17 return num; 18 } 19 int main() 20 { 21 char Binarystring[200]; 22 int N = 0; 23 printf("输入二进制字符串："); 24 gets(Binarystring); 25 N = Bin2Digit(Binarystring,strlen(Binarystring)); 26 printf("N = %d ",N); 27 return 0; 28 }
运行结果1：
代码2：
1 #include <stdio.h> 2 #include<string.h> 3 struct complex //定义复数结构体 4 { 5 float real; 6 float maginary; 7 }; 8 void PrintComplex(struct complex *member){ //打印复数 9 if(member->maginary < 0) //这里的操作是为了输出的美观 10 printf(" %f - %fj ",member->real,abs(member->maginary)); 11 else 12 printf(" %f + %fj ",member->real,member->maginary); 13 }//复数加法，返回复数结构体 14 struct complex AddComplex(struct complex *member1,struct complex *member2){ 15 struct complex child; 16 child.real = member1->real + member2->real; //实部1+实部2 17 child.maginary = member1->maginary + member2->maginary;//虚部1+虚部2 18 //printf("c1 + c2 = "); 19 //PrintComplex(&child); 20 return child; 21 } 22 //复数减法，返回复数结构体 23 struct complex SubComplex(struct complex *member1,struct complex *member2){ 24 struct complex child; 25 child.real = member1->real - member2->real; 26 child.maginary = member1->maginary - member2->maginary; 27 return child; 28 } 29 //复数乘法，返回复数结构体 30 struct complex MulComplex(struct complex *member1,struct complex *member2){ 31 struct complex child; 32 child.real = (member1->real * member2->real) - (member1->maginary * member2->maginary); 33 child.maginary = (member1->maginary * member2->real) + (member1->real * member2->maginary); 34 return child; 35 } 36 int main(int argc, char const *argv[]) 37 { 38 struct complex c1,c2,temp; 39 printf("Input c1 : "); //分别输入两个结构体的实部、虚部值 40 scanf("%f %f",&c1.real,&c1.maginary); 41 printf("Input c2 : "); 42 scanf("%f %f",&c2.real,&c2.maginary); 43 printf("c1 is:"); //打印原始的两个复数 44 PrintComplex(&c1); 45 printf("c2 is:"); 46 PrintComplex(&c2); 47 temp = AddComplex(&c1,&c2); //复数加，函数结果赋值给中间变量结构体temp 48 printf("c1 + c2 = "); 49 PrintComplex(&temp); 50 temp = SubComplex(&c1,&c2); //复数减，函数结果赋值给中间变量结构体temp 51 printf("c1 - c2 = "); 52 PrintComplex(&temp); 53 temp = MulComplex(&c1,&c2); //复数乘，函数结果赋值给中间变量结构体temp 54 printf("c1 * c2 = "); 55 PrintComplex(&temp); 56 return 0; 57 }
运行结果2：
代码3：
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 int my_atoi(char s) //数字字符转数字 5 { 6 return s-'0'; 7 } 8 9 int main(){ 10 unsigned long BigNumber=0,num=0; 11 int iArray[10]={0}; 12 char sArray[11]={0}; 13 char binBuffer[33]={0}; 14 int index=0; 15 printf("Input a big number: "); 16 scanf("%ld",&BigNumber); //现有用户输入一个数字 17 printf("BigNumber = %d ", BigNumber); //打印数字 18 itoa(BigNumber,sArray,10); // 十进制 整数转对应的十进制字符串 19 //打印 数字字符串数组 和 长度 20 printf("sArray = '%s' ,length = %d ",sArray,strlen(sArray)); 21 while( index < strlen(sArray) ) //把数字字符串挨个转成整数填入数字数组 22 { 23 iArray[index] = my_atoi(sArray[index]); 24 index += 1; 25 } 26 // 打印整数数组 27 for(index=0;index<strlen(sArray);index++) 28 printf("%d ",iArray[index]); 29 index = 0; 30 while( index < strlen(sArray) )//从整数数组累加恢复开始的整数数字 31 { 32 num *= 10; 33 num += iArray[index]; 34 index += 1; 35 } 36 //调用itoa直接得到整数的二进制字符串，参数2代表二进制 37 printf("Number from array is %d, binary string is %s.",num ,itoa(num,binBuffer,2)); 38 return 0; 39 }
运行结果3：
代码4：
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <string.h> 4 #include <math.h> 5 int Calc(int index,int len) //计算按照规则本数字的下一个数字，len为数字长度 6 { 7 int i = 0,res = 0,temp=0; //res累加其结果 8 while(i < len) 9 { 10 index %= (int)pow(10,len-i); 11 temp = index / (int)pow(10,len-i-1); 12 res += (int)pow(temp,2); 13 i += 1; 14 } 15 return res; 16 } 17 int print_link(int num,int len) //打印单行列表，num为初始数字，len为其长度 18 { 19 char buf[10]; 20 itoa(num,buf,10); //这一步是为了后面求出数字长度 21 while(num > 9) //打印终止于小于10的数字 22 { 23 printf("%d -->",num); 24 num = Calc(num,strlen(buf)); 25 memset(buf,0,sizeof(buf)); 26 itoa(num,buf,10); 27 } 28 printf("%d ", num); 29 return num; 30 } 31 int main(){ 32 unsigned int number = 0,temp = 0; 33 char NumBuffer[10]={0}; 34 printf("Input a number: "); 35 gets(NumBuffer); //读取字符串形式整数 36 temp = atoi(NumBuffer); //转成整数 37 while(temp>=10) 38 { 39 if(1 ==print_link(temp,strlen(NumBuffer))) //记录结果为1的数目 40 number++;; 41 temp -= 1; 42 } 43 printf("Number of '1' is %d ",number); 44 return 0; 45 }
运行结果4：
