JOJ的2042題目是一個(gè)程序理解題目，這個(gè)題目非常有意思，給出了下面一段C++源代碼，要求計(jì)算出最后的輸出結(jié)果，源代碼如下：

#include<cstdio>
int main(void)
{
???? int x = 987654321, c = 0, d = 1, e = 6;
???? while(x--){
???????? c += d,
???????? d += e,
???????? e += 6;
???? }
???? printf("%d/n", c);
???? return 0;
}

原題目如下：

We can use axioms to calculate programs just like what we do in algebras. Dijkstra is the one who advocates such constructive approach whereby a program is designed together with its correctness proof. In short, one has to start from a given postcondition Q and then look for a program that establishes Q from the precondition. Often, analyzing Q provides interesting hints to finding the program. This approach is quite different from the well known Hoare Logic.

?

For example, the following program is calculated by Dijkstra's approach. Unfortunately, its annotation is lost so that its function is hard to grasp. You are to help with finding the final value of the variable c. Note that the program is designed under 128-bit architecture.

代碼就是上面那一段。

這個(gè)題目通過小數(shù)據(jù)計(jì)算可以看出規(guī)律：x=1, c = 1; x=2, c=8; x=3, c=27; x=4, c=64，于是可以猜測這段程序是用來計(jì)算x^3的。用計(jì)算器計(jì)算出987654321^3，提交上去就AC了。

這個(gè)題目是超級大牛SIYEE出的。但是為什么會(huì)得到這樣神奇的結(jié)果？有數(shù)學(xué)推導(dǎo)方法嗎？

有！而且是高中數(shù)學(xué)知識!這時(shí)候才知道高中數(shù)學(xué)是多么有用呵！

首先看這個(gè)程序：

int x = 987654321, c = 0, d = 1, e = 6;
???? while(x--){
???????? c += d,
???????? d += e,
???????? e += 6;
???? }

這里c,d,e充當(dāng)了一個(gè)臨時(shí)寄存器角色，不要被賦值語句迷惑了，其實(shí)c,d,e都是數(shù)列

首先在草稿紙上列出前幾列：

x??0???1??? 2

c? 0?? 1??? 8

d ?1?? 7?? 19

e? 6?? 12? 18

?

然后在C語言描述中看：

（1）c的數(shù)列描述：

c(n) = 0;??? n = 0

c(n) = c(n -1) + d(n - 1); n>=1

（2）d的數(shù)列描述:

d(n) = 1;??? n = 0

d(n) = d(n -1) + e(n - 1); n>=1

（3）e的數(shù)列描述：

e(n) = 6(n+1)

由于e(n)已知，所以可以從這里入手：

計(jì)算d(n):

d(n) = d(n -1) + e(n - 1); n>=1

變換：

d(n) - d(n -1)?= e(n - 1) = 6n

可以看出是一個(gè)等變數(shù)列，用累加法：

d(n) - d(n -1)?=? 6n

d(n -1) - d(n -2)?=? 6(n-1)

...............................

d(1) - d(0) = 6

累加：

d(n) - d(0) = 6n+6(n-1)+........+6(1) = 3n(n+1)

d(n) = 3n(n+1)+1

驗(yàn)算一下，沒錯(cuò)，那么繼續(xù)求c(n):

用同樣方法 ：

c(n) = c(n -1) + d(n - 1);

c(n) - c(n-1) = 3(n -1)n +1 = 3(n^2) -3n +1

c(n-1) - c(n-2) = 3(n -2)(n-1) +1 = 3((n-1)^2) -3(n-1) +1

.................................................

c(1) - c(0) = 3(1^2) - 3(1) +1

累加：

c(n)-c(0) = 3[n^2+(n-1)^2+.......+2^2+1^2]-3[n+(n-1)+...+2+1]+n

c(n)-c(0) = 1/2[2(n)^3+3(n)^2+n]-3/2[(n)^2+n]+n

c(n)-c(0) = (n)^3+3/2(n)^2+1/2 *(n)-3/2 *(n^2)-3/2 *n +n

c(n)-c(0) = n^3,已知 C(0) = 0

故得c(n) = n^3?? n = 0,1,2.......

總結(jié)

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