范围内整数的二进制补码二进制表示中 1 的数量 [英] Number of 1s in the two's complement binary representations of integers in a range

问题描述

这个问题来自 2011 Codesprint (http://csfall11.interviewstreet.com/):

This problem is from the 2011 Codesprint (http://csfall11.interviewstreet.com/):

计算机科学的基础之一是了解数字在 2 的补码中的表示方式.想象一下，您使用 32 位以 2 的补码表示形式写下 A 和 B 之间的所有数字.你一共要写多少个1?输入:第一行包含测试用例的数量 T (<1000).接下来的 T 行中的每一行都包含两个整数 A 和 B.输出:输出 T 行，一个对应每个测试用例.约束:-2^31 <= A <= B <= 2^31 - 1

One of the basics of Computer Science is knowing how numbers are represented in 2's complement. Imagine that you write down all numbers between A and B inclusive in 2's complement representation using 32 bits. How many 1's will you write down in all ? Input: The first line contains the number of test cases T (<1000). Each of the next T lines contains two integers A and B. Output: Output T lines, one corresponding to each test case. Constraints: -2^31 <= A <= B <= 2^31 - 1

示例输入:3-2 0-3 4-1 4样本输出:639937

Sample Input: 3 -2 0 -3 4 -1 4 Sample Output: 63 99 37

说明:对于第一种情况，-2 包含 31 个 1，后跟一个 0，-1 包含 32 个 1，0 包含 0 个 1.因此总数为 63.对于第二种情况，答案是 31 + 31 + 32 + 0 + 1 + 1 + 2 + 1 = 99

Explanation: For the first case, -2 contains 31 1's followed by a 0, -1 contains 32 1's and 0 contains 0 1's. Thus the total is 63. For the second case, the answer is 31 + 31 + 32 + 0 + 1 + 1 + 2 + 1 = 99

我意识到您可以利用 -X 中 1 的数量等于 (-X) = X-1 的补码中 0 的数量这一事实来加快搜索速度.该解决方案声称存在用于生成答案的 O(log X) 递归关系，但我不明白.解决方案代码可以看这里:https://gist.github.com/1285119

I realize that you can use the fact that the number of 1s in -X is equal to the number of 0s in the complement of (-X) = X-1 to speed up the search. The solution claims that there is a O(log X) recurrence relation for generating the answer but I do not understand it. The solution code can be viewed here: https://gist.github.com/1285119

如果有人能解释这种关系是如何得出的，我将不胜感激！

I would appreciate it if someone could explain how this relation is derived!

推荐答案

嗯，没那么复杂...

单参数solve(int a) 函数是关键.它很短，所以我将在此处剪切和粘贴:

The single-argument solve(int a) function is the key. It is short, so I will cut&paste it here:

long long solve(int a)
{
 if(a == 0) return 0 ;
 if(a % 2 == 0) return solve(a - 1) + __builtin_popcount(a) ;
 return ((long long)a + 1) / 2 + 2 * solve(a / 2) ;
}

它只对非负a有效，它计算从0到a(含)所有整数中1的位数.

It only works for non-negative a, and it counts the number of 1 bits in all integers from 0 to a inclusive.

函数分三种情况:

a == 0 -> 返回 0.显然.

a == 0 -> returns 0. Obviously.

a even -> 返回 a 中 1 的位数加上 solve(a-1).也很明显.

a even -> returns the number of 1 bits in a plus solve(a-1). Also pretty obvious.

最后一个案例很有趣.那么，我们如何计算从 0 到奇数 a 的 1 位数呢?

The final case is the interesting one. So, how do we count the number of 1 bits from 0 to an odd number a?

考虑 0 到 a 之间的所有整数，并将它们分成两组:偶数和赔率.例如，如果 a 为 5，则您有两个组(二进制):

Consider all of the integers between 0 and a, and split them into two groups: The evens, and the odds. For example, if a is 5, you have two groups (in binary):

000  (aka. 0)
010  (aka. 2)
100  (aka. 4)

和

001  (aka 1)
011  (aka 3)
101  (aka 5)

注意这两个组必须具有相同的大小(因为 a 是奇数，并且范围包括在内).要计算每组中有多少个 1，首先数除最后一位之外的所有位，然后数最后一位.

Observe that these two groups must have the same size (because a is odd and the range is inclusive). To count how many 1 bits there are in each group, first count all but the last bits, then count the last bits.

除了最后几位之外的所有内容如下所示:

All but the last bits looks like this:

00
01
10

...both 组看起来像这样.这里 1 的位数就是 solve(a/2).(在这个例子中，它是从 0 到 2 的 1 的位数.另外，请回忆一下 C/C++ 中的整数除法会将 向下舍入.)

...and it looks like this for both groups. The number of 1 bits here is just solve(a/2). (In this example, it is the number of 1 bits from 0 to 2. Also, recall that integer division in C/C++ rounds down.)

对于第一组中的每个数字，最后一位为零，对于第二组中的每个数字，最后一位都为零，因此最后一位在总数中贡献了 (a+1)/2 一位.

The last bit is zero for every number in the first group and one for every number in the second group, so those last bits contribute (a+1)/2 one bits to the total.

所以递归的第三种情况是(a+1)/2 + 2*solve(a/2)，适当的转换为long long来处理a 是 INT_MAX 的情况(因此 a+1 溢出).

So the third case of the recursion is (a+1)/2 + 2*solve(a/2), with appropriate casts to long long to handle the case where a is INT_MAX (and thus a+1 overflows).

这是一个 O(log N) 的解决方案.要将其推广到 solve(a,b)，您只需计算 solve(b) -solve(a)，加上担心负数的适当逻辑.这就是两个参数 solve(int a, int b) 正在做的事情.

This is an O(log N) solution. To generalize it to solve(a,b), you just compute solve(b) - solve(a), plus the appropriate logic for worrying about negative numbers. That is what the two-argument solve(int a, int b) is doing.

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