华科14_2

问题：

以二进制方式输入两个正整数，然后再输入一个四则运算符号，按short类型计算，并将结果按二进制输入。

分析：

自己写二进制的四则运算觉得比较麻烦，干脆通过十进制，先将要计算的数转换为10进制，然后进行相应计算，然后将结果再转换为二进制进行输出。

代码：

#include <stdio.h>   #include <string.h>  #include <math.h>  #include <vector>  #include <queue>  #include <stack>  #include <map>  #include <string>  #include <algorithm>  #include <iomanip>#define MAX 1000using namespace std;/* run this program using the console pauser or add your own getch, system("pause") or input loop */struct Binary{char num[MAX];void init(char *str){strcpy(num,str);}void init(int x){//用10进制数x 初始化二进制数 int size=0;while(x != 0) {num[size++] = (x%2) + '0';x/=2;}num[size++] = '\0';}int toTen()const{//将二进制数b 转化为10进制数ansint ans=0;int c=1;int length = strlen(num);for(int i=length-1;i>=0;i--){ans+=(num[i]-'0')*c;c*=2;}return ans;}Binary operator *(const Binary &b)const{int m=this->toTen();int n=b.toTen();Binary ans;ans.init(m*n);return  ans;}Binary operator /(const Binary &b)const{int m=this->toTen();int n=b.toTen();Binary ans;ans.init(m/n);return  ans;}Binary operator +(const Binary &b)const{int m=this->toTen();int n=b.toTen();Binary ans;ans.init(m+n);return  ans;}Binary operator -(const Binary &b)const{int m=this->toTen();int n=b.toTen();Binary ans;ans.init(m-n);return  ans;}void  display(){int length = strlen(num);for(int i=length-1;i>=0;i--){printf("%c",num[i]);}printf("\n");}};int main(int argc, char** argv) {char num1[MAX],num2[MAX];char op;while(scanf("%s %s %c",num1,num2,&op)!=EOF){Binary a,b;a.init(num1);b.init(num2);if(op == '+'){Binary c= a+b;c.display();}else if(op == '-'){Binary c= a-b;c.display();}else if(op == '*'){Binary c= a*b;c.display();}else if(op == '/'){Binary c= a/b;c.display();}}return 0;}