描述

将一个长度最多为30位数字的十进制非负整数转换为二进制数输出。

输入

多组数据，每行为一个长度不超过30位的十进制非负整数。

（注意是10进制数字的个数可能有30个，而非30bits的整数）

输出

每行输出对应的二进制数。

样例输入

0

1

3

8

样例输出

0

1

11

1000

分析问题：

将一个长度最多为30个数字的十进制非负整数转换为二进制数输出。这是大数问题，肯定需要用字符串来存储这个十进制数，转化出来的二进制也需要用字符串来存储。和小数的转化一样，不断取2的模，然后除2。

思路：

字符串模2除2

可以优化程序。//冰非寒(binfeihan@126.com) //252kB 30ms 1096 B G++ #include <iostream> #include <string> using namespace std; const int N = 31; //class: Binary typedef class Binary{ public: void dec2bin(string s); void print(); private: string BinaryNum; }Binary; //convert decimal to Binary void Binary::dec2bin(string s){ int sum(0),j(0); string bin; int an[N]={0}; //assign to an for(int i(s.length() – 1);i >= 0;– i){ an[j] = s[i] – ‘0’; sum += an[j]; ++ j; } //calculate the Binary number while(1){ char ch = an[0] % 2 + ‘0’; bin.insert(bin.length(),1,ch); sum = 0; for(int i(j – 1);i >= 0;– i){ if(i > 0){ an[i – 1] += an[i] % 2 * 10; } an[i] /= 2; sum += an[i]; } if(sum == 0){ break; } } //reverse the Binary number and assign for(int i(bin.length() – 1);i >= 0;– i){ this->BinaryNum.insert(this->BinaryNum.length(),1,bin[i]); } } //print void Binary::print(){ //print the Binary number cout<<this->BinaryNum<<endl; } //main function int main(){ string dec; while(cin>>dec){ Binary b; b.dec2bin(dec); b.print(); } return 0; }

改进后的代码：
时间分析
1.cout,cin读写比scanf,printf慢
2.调用string函数比直接操作char型字符数组占时间

//冰非寒(binfeihan@126.com) //256kB 0ms 1364 B G++ #include <stdio.h> #include <string.h> const int N = 31; const int M = 200; //class: Binary typedef class Binary{ public: void dec2bin(char s[]); void print(); private: char BinaryNum[M]; }Binary; //convert decimal to Binary void Binary::dec2bin(char s[]){ int i(0),sum(0),j(0),k(0),t(1),p(0); char ch; char bin[M]; int an[N]={0}; //assign to an for(i = strlen(s) – 1;i >= 0;– i){ if(k < 4){ an[j] = an[j] + (s[i] – ‘0’) * t; t *= 10; k ++; } if(k == 4){ k = 0; ++ j; t = 1; continue; } } if(k != 0){ ++ j; } //calculate the Binary number k = 0; while(1){ ch = an[0] % 2 + ‘0’; bin[p ++] = ch; sum = 0; for(i = j – 1;i >= 0;– i){ if(i != 0){ an[i – 1] += an[i] % 2 * 10000; } an[i] /= 2; sum += an[i]; if(k == 0 && an[i] > 0){ t = i + 1; k = 1; } } if(sum == 0){ break; } j = t; k = 0; } //reverse the Binary number and assign j = 0; for(i = p – 1;i >= 0;– i){ this->BinaryNum[j ++] = bin[i]; } this->BinaryNum[j] = ‘\0’; } //print void Binary::print(){ //print the Binary number printf(“%s\n”,this->BinaryNum); } //main function int main(){ char dec[M]; while(scanf(“%s”,dec) == 1){ Binary b; b.dec2bin(dec); b.print(); } return 0; }