CodeForces 9C -Hexadecimal's Numbers-构造

问题描述：

One beautiful July morning a terrible thing happened in Mainframe: a mean virus Megabyte somehow got access to the memory of his not less mean sister Hexadecimal. He loaded there a huge amount of n different natural numbers from 1 to n to obtain total control over her energy.

But his plan failed. The reason for this was very simple: Hexadecimal didn’t perceive any information, apart from numbers written in binary format. This means that if a number in a decimal representation contained characters apart from 0 and 1, it was not stored in the memory. Now Megabyte wants to know, how many numbers were loaded successfully.

输入：

Input data contains the only number n (1 ≤ n ≤ 109).

输出：

Output the only number — answer to the problem.

样例：

Input

10

Output

2

题目大意：

一个整数n，求1到n这n个数中有几个数是由1和0组成的

AC代码：

int bit[100];//用整形数组存放二进制数
char bitt[100];
int main()
{
int n,i,j,sum=1,len;
scanf("%d",&n);//输入
bit[0]=1;//初始化，从1开始，bit[0]储存这个二进制数的位数
bit[1]=1;
while(1)
{
bit[1]++;//给这个二进制数加1
sum++;//情况数加1
//维护这个二进制数
for(i=1;i<=bit[0];i++)
if(bit[i]==2)
{
bit[i]=0;
bit[i+1]++;
}
else
break;
//求出当前二进制数的位数
for(i=20;i>=1;i--)//倒着循环
if(bit[i]==1)
{
bit[0]=i;
break;
}
len=int(log10(n))+1;//len为十进制数n的位数
j=0;//数组下标
//如果二进制的位数大于了n的位数-1，就开始判断是否得出了解
if(bit[0]>=len-1)
{
for(i=bit[0];i>=1;i--)//将二进制数从int数组转为char数组,因为需要使用strtol函数
{
bitt[j]=bit[i]+'0';
j++;
}
if(strtol(bitt,NULL,10)>n)//如果当前的二进制数大于了n
{
sum--;//情况数减1
printf("%d\n",sum);//打印
return 0;
}
}
}
}

解决方法：

由0和1想到二进制，于是构造一个二进制数，如10001来代表十进制数一万零一，从一开始，这个二进制数每加1所得到的新二进制数都是满足题意的。

又想到利用高精度的思想去实现这个二进制数。

PS:在C语言某个程序当中需要把文本16进制转换成对应的16进制数，比如字符串”0x1a”转换成10进制的26，可以用以下函数来实现

相关函数: atof, atoi, atol, strtod, strtoul

表头文件: #include

int main()
{
char a[] = "100";
char b[] = "100";
char c[] = "0x11";
int x, y, z;
x = strtol( a, NULL, 10 );
y = strtol( b, NULL, 2 );
z = strtol( c, NULL, 16 );
printf( "x = %d\n", x );
printf( "y = %d\n", y );
printf( "z = %d\n", z );
}

输出：

x = 100

y = 4

z = 17