进制转换

1.二进制转十进制

#include  <iostream>
#include  <string.h>
void main()
{
	long int i,len,sum=0;
	char str[30];
	printf("\n输入二进制数:\n");
	gets(str);
	len=strlen(str);
	for(i=len-1;i>=0;i--)
		sum+=(long)(str[i]-'0')<<(len-1-i);
	printf("%ld\n",sum);
	system("pause");
} 

2.获得多少个1

#include <iostream>
using namespace std;
char *ok(int n,char *b)  
{  
	static int LEN=8*sizeof(int);  
	for(int i=LEN-1;i>=0;i--,n>>=1)  
		b[i]=(01&n)+'0';  
	b[LEN]='\0';  
	return b;  
}  

int GetIntCount(string b)
{
	int nCount=0;
	static int LEN=8*sizeof(int); 
	for(int i=0;i<LEN;i++)
	{
		if(1==b[i]-48)
			nCount++;
	}
	return nCount;
}

void main()  
{  
	int v=255;  
	char b[8*sizeof(int)+1];  
	int i=-1;  
	string s=ok(v,b);
	cout<<ok(v,b)<<endl;
	cout<<GetIntCount(s)<<endl;
	system("pause");
} 

#include <iostream>
using namespace std;

int GetIntCount(int n)
{
	char Answer[8*sizeof(int)+1];  
	static int LEN=8*sizeof(int);
	for (int i = LEN - 1; i >= 0; i--, n >>= 1)
		Answer[i] = (01 & n) + '0';
	Answer[LEN] = '\0';
	string Str = Answer;
	int nCount=0;
	for(int i=0;i<LEN;i++)
	{
		if(1==Str[i]-48)
			nCount++;
	}
	return nCount;
}
void main()  
{  
	cout<<GetIntCount(7)<<endl;
	system("pause");
} 

3.十进制转二进制

#include <iostream>
using namespace std;
char *ok(int n,char *b)  
{  
	static int LEN=8*sizeof(int);  
	for(int i=LEN-1;i>=0;i--,n>>=1)  
		b[i]=(01&n)+'0';  
	b[LEN]='\0';  
	return b;  
}  
void main()  
{  
	int v[]={7,256};  
	char b[8*sizeof(int)+1];  
	int i=-1;  
	while(++i<sizeof(v)/sizeof(v[0]))  
		cout<<ok(v[i],b)<<endl;
	system("pause");
} 

#include <stdlib.h> 
#include <stdio.h> 

int Fuc(int bin)
{
	int lln=1;
	int dec=0 ;
	while (bin)
	{
		dec+=bin%10*lln;
		lln*=2;
		bin/=10;
	}
	return dec;
} 
int main(void) 
{    
	int number = 256;     
	char string[25];   
	printf("%d\n",number);
	itoa(number, string, 2);  
	printf("%s\n",string);   
	number = atoi(string);  
	printf("%d\n",Fuc(number));
	system("pause");
	return 0;
}