CarceerCup 系列 第五章 Bit Manipulation

题目翻译和代码有不正确的地方欢迎批评指正.有好的方法欢迎交流。

第一题：

You are given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method to set all bits between i and j in N equal to M (e.g., M becomes a substring of N located at i and starting at j).
EXAMPLE:
Input: N = 10000000000, M = 10101, i = 2, j = 6
Output: N = 10001010100

把N从左边数的第i到j位设置成M的值

代码如下：

#include<stdio.h>

int main()
{
	int N, M;
	int i, j;
	int x = 0;
	scanf("%d %d", &N, &M);
	scanf("%d %d", &i, &j);
	x = (0|M)<<i; //000M00
	N &= (((~0) << (j-i+1)) <<i) | (~((~0)<<i)); //N000N
	N |= x;
	printf("%d %d", N, M);
}

第二题：

Given a (decimal - e.g. 3.72) number that is passed in as a string, print the binary representation. If the number can not be represented accurately in binary, print “ERROR”

还不知道怎么写代码。

第三题：

Given an integer, print the next smallest and next largest number that have the same number of 1 bits in their binary representation.

给你一个整数n，求出最小的比n大并且和n的二进制中1的个数一样多的数。找出最大的比n小并且和n含的1一样多的数。

方法一：

从n开始依次递增，递减，找到满足条件的数。

方法二：

寻找大数：

1.从n的二进制数的右边往左寻找到第一位是1，并且该位左边的一位是0

2.把找到的这一位变为0，左边的变为1；

3.把变为0的那一位右边的1全部移到最右边（最低位）

寻找小数原理相似。

代码如下：

#include<stdio.h>
int Findlarge(int);
int Findsmall(int);
int main()
{
	int n;
	int small = 0, large = 0;
	scanf("%d", &n);
	assert(n >1)
	large = Findlarge(n);
	small = Findsmall(n);
	printf("%d %d", large, small);
}

int Findlarge(int n)
{
	int k, count;
	k = 1;
	count = 0;
	while((n & k) == 0)//跳过开始的0
	{
		k<<= 1;
	}
	while((n&k)!= 0)//寻找位的值为1，但是右边的位为0 的位
	{
		k<<= 1;
		count++;
	}
	n |= k;//
	k >>= 1;
	n&= ~k;
	count--; //右边1的个数，
    while(k > 0)
	{
		if((n &k)!= 0)
			n &= ~k;
		k >>= 1;
	}
	n |= ~((~0)<<count);
	return n;
}

int Findsmall(int n)
{
	int k, count;
	k = 1;
	count = 0;
	while((n&k) != 0)
	{
		k<<= 1;
		count++;
	}
	while((n&k) == 0)
	{
		k<<= 1;
	}
	n &= ~k;
	k >>= 1;
	n |= k;
	while(k > 0)
	{
		k >>= 1;
		if(count > 0)
			n |= k;
		else
		{
			n &= ~k;
		}
		count--;
	}
	return n;
}

第四题：

Explain what the following code does: ((n & (n-1)) == 0).

用来判断n的二进制中是不是只含有一个1，

第五题：

Write a function to determine the number of bits required to convert integer A to integer B.
Input: 31, 14
Output: 2

其实也就是判断两个数，有多少位不同。

用上一题的结论。

代码如下：

#include<stdio.h>

int main()
{
	int n, m;
	int count;
	scanf("%d %d", &n, &m);
	m ^= n;
	count = 0;
	while(m)
	{
		count++;
		m &= m-1;
	}
	printf("%d\n", count);
}

第六题：

Write a program to swap odd and even bits in an integer with as few instructions as possible (e.g., bit 0 and bit 1 are swapped, bit 2 and bit 3 are swapped, etc).

交换一个数字的奇数位和偶数位.

代码如下：

#include<stdio.h>
int main()
{ 
	int x;
	scanf("%d", &x);
	x = ((x& 0xaaaaaaaa)>>1) | ((x&0x55555555)<<1);
	printf("%d", x);
}

第七题：

An array A[1... n] contains all the integers from 0 to n except for one number which is missing. In this problem, we cannot access an entire integer in A with a single operation. The elements of A are represented in binary, and the only operation we can use to access them is “fetch the jth bit of A[i]”, which takes constant time. Write code to find the missing integer. Can you do it in O(n) time?

题意如下：

给你一个数组A[n]，里面存储了0到n的n个数字，你不能获得数组的数字，只能知道数组中数字的第k位是1还是0，你能在o(N)的时间里找到不在数组中的那个数字吗？

思路如下：

先求出n的有效位数，假如为k位，创建一个一个大小为2^k+1次方大小的数组，复制A[n]中的元素到这个数组中，后面的元素从n依次加1；

然后统计每个位上面1和0的个数，就可以知道要找的数字的每一位是0还是1.

假如n=4，扩充到7，使每一位上面的0和1的个数想同。

0000

0001

0010

0011

0100

0101

0110

0111

代码如下：

#include<stdio.h>
#include<malloc.h>
#include<assert.h>

void Find(int* ,int , int , int *);
int GetBitof1(int *, int , int);//获得第k位0或者1的个数
int main()
{
	int *A, n;
	int i;
	int result = 0;
	int m = 1;
	int count = 0;
	int temp;
	scanf("%d", &n);
	assert(n > 0);
	temp = n;
	while(temp)//统计n的有效位
	{
		temp >>= 1;
		count++;
	}
	//count++;
	m <<= count;
	A = (int *)malloc(sizeof(int) * (m-1));
	assert(A != NULL);
	printf("Input the value form 0~n\n");
	for(i = 0;i < n; ++i)
		scanf("%d", A+i);
	for(; i < m-1; ++i)//扩充数组，使第k位上面1和0的个数相同
		A[i] = i+1;

	for(i = 0; i < m-1; ++i)//观察结果
		printf("%d ", A[i]);
	printf("\n");

	Find(A, m-1, 1, &result);
	printf("%d ", result);
	free(A);
	return 0;
}

void Find(int *A, int n, int k, int *result)
{
	int m = 1;
	int count1 = 0;
	int count0 = 0;
	m <<= k-1;
	if(m > n)
		return;
	count1 = GetBitof1(A, n, k);
	count0 = n-count1;
	if(count0 > count1)//0比1多，说明这一位应该是1
	{
		*result |= m;
	//	Find(A, n, k+1, result);
	}
	Find(A, n, k+1, result);
}

int GetBitof1(int *A, int n, int k)
{
	int count = 0;
	int i;
	int value = 1;
	value <<= k-1;
	for(i = 0; i < n; ++i)
	{
		if(A[i] &value)
			count++;
	}
	return count;
}