外国人的数学素质是差啊

caimicn · 发表于 2016-3-31 09:14

游戏里有这样的谜题：要求
（A+B-C)*D+E-F=7
A、B、C、D、E、F为11、9、1、6、4、2的其中一个数值。
好吧！！
第一解：

（9-6+1）*4+2-11=7

居然没通过

好吧！继续，第二解：

（9-4+1）*2+6-11=7

又说我是错的，

没关系，我还有第三解：

（11-6+1）*2+4-9=7

我靠，又说我错的，我CAO，老外的数学全都是体育老师教的吗？我崩溃了

还有没有第4解啊？第5解啊？？

caimicn · 发表于 2016-3-31 10:14

好了，找到第四种方法了

myq495 · 发表于 2016-3-31 11:06

国内人数学也有差的，不能以偏概全

xuyiqing · 发表于 2016-3-31 22:55

哪个游戏啊？

飞翔的荷兰人 · 发表于 2016-4-8 16:03

如果老毛子不算外国人的话，楼主的话勉强算正确。

iguodala · 发表于 2016-4-30 11:08

老外的数学绝对没我们学的深

tianzhe · 发表于 2016-5-1 23:45

有一次有个游戏居然问X*Y=63，X和Y是多少，会背九九表的人都知道，可见他们的数学基础

血腥的情人节 · 发表于 2016-6-23 11:23

代入以下代码，得：11,9,4,2,1,6

滑稽~

#include <iostream>
#include <string.h>
using namespace std;
template <class T>
class Node
{
public:
T data;
Node<T> *next;
Node()
{
this->next = NULL;
}
Node(T data,Node<T> *next=NULL)
{
this->data = data;
this->next = next;
}
};

template <class T>
class LinkedStack
{
private:
Node<T> *top;
public:
LinkedStack();
~LinkedStack();
bool isEmpty();
void push(T x);
T pop();
T get();
};

template <class T>
LinkedStack<T>::LinkedStack()
{
top = NULL;
}

template <class T>
LinkedStack<T>::~LinkedStack()
{
Node<T> *p = top;
Node<T> *q;
while(p!=top)
{
q = p;
p = p->next;
delete p;
}
top = NULL;
}

template <class T>
bool LinkedStack<T>::isEmpty()
{
return top == NULL;
}

template <class T>
void LinkedStack<T>::push(T x)
{
top = new Node<T>(x,top);
}

template <class T>
T LinkedStack<T>::pop()
{
if(!isEmpty())
{
T x = top->data;
Node<T> *p = top;
top = top->next;
delete p;
return x;
}
throw "空栈，不能执行出栈操作";
}

template <class T>
T LinkedStack<T>::get()
{
if(!isEmpty())
{
return top->data;
}
throw "空栈，不能获得栈顶元素";
}

char * toPostfix(char *expstr)
{
LinkedStack<char> stack;
char *postfix = new char[strlen(expstr)*2];
int i=0;
int j=0;
char out ;
while(expstr[i]!='\0')
{
switch(expstr[i])
{
case'+':
case'-':
while(!stack.isEmpty()&&stack.get()!='(')
{
postfix[j++] = stack.pop();
}
stack.push(expstr[i++]);
break;
case'*':
case'/':
while(!stack.isEmpty()&&(stack.get()=='*'||stack.get()=='/'))
{
postfix[j++] = stack.pop();
}
stack.push(expstr[i++]);
break;
case'(':stack.push(expstr[i++]);
break;
case')':out = stack.pop();
while(!stack.isEmpty()&&out!='(')
{
postfix[j++] = out;
out = stack.pop();
}
i++;
break;
default:
while(expstr[i]>='0'&&expstr[i]<='9'&&expstr[i]!='\0')
{
postfix[j++] = expstr[i++];
}
postfix[j++]=' ';
break;
}
}
while(!stack.isEmpty())
{
postfix[j++]=stack.pop();
}
postfix[j]='\0';
return postfix;
}

int value(char *postfix)
{
LinkedStack<int> stack;
int i=0;
int result = 0;
while(postfix[i]!='\0')
{
if(postfix[i]>='0'&&postfix[i]<='9')
{
result = 0;
while(postfix[i]!=' ')
{
result = result*10+postfix[i++]-'0';
}
i++;
stack.push(result);
}
else
{
if(postfix[i]!=' ')
{
int y = stack.pop();
int x = stack.pop();
switch(postfix[i])
{
case'+':result = x + y;
break;
case'-':result = x - y;
break;
case'*':result = x *y;
break;
case'/':result = x / y;
break;
}
stack.push(result);
}
i++;
}
}
return stack.pop();
}

int main()
{
//char *expstr = "121+10*(52-49+20)/((35-25)*2+10)";
cout << "请输入表达式:";
//char *a ;
//cin >> *a;
char expstr[20]={0};
while(1)
{
cin>>expstr;
char *postfix = toPostfix(expstr);
cout << "expstr= "<<expstr << endl;
cout << "postfix= "<<postfix<<endl;
cout << "value= "<<value(postfix) << endl;
}
return 0;
}

血腥的情人节 · 发表于 2016-6-23 11:24

caimicn 发表于 2016-3-31 10:14
9 q3 [ Q$ B" T好了，找到第四种方法了

代入以下代码，得：11,9,4,2,1,6

滑稽~

#include <iostream>
#include <string.h>
using namespace std;
template <class T>
class Node
{
public:
T data;
Node<T> *next;
Node()
{
this->next = NULL;
}
Node(T data,Node<T> *next=NULL)
{
this->data = data;
this->next = next;
}
};

template <class T>
class LinkedStack
{
private:
Node<T> *top;
public:
LinkedStack();
~LinkedStack();
bool isEmpty();
void push(T x);
T pop();
T get();
};

template <class T>
LinkedStack<T>::LinkedStack()
{
top = NULL;
}

template <class T>
LinkedStack<T>::~LinkedStack()
{
Node<T> *p = top;
Node<T> *q;
while(p!=top)
{
q = p;
p = p->next;
delete p;
}
top = NULL;
}

template <class T>
bool LinkedStack<T>::isEmpty()
{
return top == NULL;
}

template <class T>
void LinkedStack<T>::push(T x)
{
top = new Node<T>(x,top);
}

template <class T>
T LinkedStack<T>::pop()
{
if(!isEmpty())
{
T x = top->data;
Node<T> *p = top;
top = top->next;
delete p;
return x;
}
throw "空栈，不能执行出栈操作";
}

template <class T>
T LinkedStack<T>::get()
{
if(!isEmpty())
{
return top->data;
}
throw "空栈，不能获得栈顶元素";
}

char * toPostfix(char *expstr)
{
LinkedStack<char> stack;
char *postfix = new char[strlen(expstr)*2];
int i=0;
int j=0;
char out ;
while(expstr!='\0')
# y" @$ H' |: f- A. s* q {
1 z. ~$ x; r) Q6 F/ @, z. f1 _ switch(expstr)! J$ Z1 ?3 N  v- C, e9 r0 ^  ~
{9 z3 }* |0 h- n
case'+':5 \- `4 X3 N0 t2 O0 }6 r0 U; L8 }7 h- p
case'-':5 W5 Z: H" \/ B
while(!stack.isEmpty()&&stack.get()!='(')& P* J% E: ^  L
{
. y# X  w" d3 m! A( W/ H postfix[j++] = stack.pop();
. B0 ~6 |, u% D- D }! H' Q5 t9 {4 D
stack.push(expstr[i++]);& W7 s2 f0 u& H% R* b
break;
5 i8 z/ G* T- u( n1 ? case'*':
/ g5 J3 u) }4 i7 o; D  r5 X case'/':4 r4 X' i* }; m  y, R+ ]" G3 P
while(!stack.isEmpty()&&(stack.get()=='*'||stack.get()=='/'))+ K4 K  c& `& Y
{
3 R( y. y% W9 H2 b, B: t- } postfix[j++] = stack.pop();1 A' d8 w- ]2 ~1 q# P" V% `
}
; ?+ \8 K- b6 u0 H- o- G stack.push(expstr[i++]);
$ J' w/ R: {* o; t$ ? break;, g" l; h, v" l; \" F+ z
case'(':stack.push(expstr[i++]);
6 y& z( x: a% j' e break;
0 B0 z, |# p0 w  s0 d2 m case')':out = stack.pop();( C) }$ A# {  G- T( v4 D
while(!stack.isEmpty()&&out!='(')4 ^$ ]& [- e. m# u5 M" U3 _) h7 k
{
/ X9 u5 r3 n; c9 f) V/ y# R9 r postfix[j++] = out;& \0 e0 X6 M' @" f
out = stack.pop();: a8 `6 U3 i  r+ t
}) c' b, v( G9 W; A, ]; [7 Y" r1 x
i++;4 l5 g/ W2 v  h6 l" o( c
break;8 N% l" ?5 X4 s1 c. s; L; k
default:
) O# X8 N' R, F  x1 g, J5 _ while(expstr>='0'&&expstr<='9'&&expstr!='\0')
# y4 m4 q8 ]! n+ t; s0 |* t {
- ~# L7 n3 v/ Q% x. D9 p5 f postfix[j++] = expstr[i++];
' n7 m8 ^0 w$ c/ D3 V }7 I; F+ {3 A3 {- k. J  L1 l
postfix[j++]=' ';, N$ A) |# ?8 F' s# N' x
break;0 K6 `5 p, `" l  \8 c% S# m
}
1 X$ ]9 a; f3 q  q7 t3 Q }! u; M/ C0 N7 p' V7 n7 x7 |
while(!stack.isEmpty())
1 g3 k& K% ?3 J* T {
* [3 m( r( O, i( v/ g8 X% W postfix[j++]=stack.pop();2 D6 J4 u% y0 q0 o/ I' u
}
) u+ Z; C" d  w postfix[j]='\0';
0 [* ~3 S3 a" H& D' ~7 c return postfix;
) n0 H, b- b( ~% |& n5 R* f1 M# C}6 |+ r6 h7 u2 F, o% j. r
1 Y# O' l& q* C
int value(char *postfix)
" r" m$ E4 @4 u* ~0 G{$ R7 N, c9 D' K! D+ _! x; u. J* o
LinkedStack<int> stack;
' ~. f# }! @4 m6 z int i=0;& I" |3 n' j$ B! N* H' s9 P1 g
int result = 0;: q/ i1 `! g5 T3 A
while(postfix!='\0')1 B2 h1 b3 O: R+ K6 n
{) _1 A; T. i$ U$ J- B: R. s
if(postfix>='0'&&postfix<='9')
6 Z; l' S4 L  F3 U* Z+ p {7 P& m) [6 a1 C4 O  R$ Q
result = 0;2 R2 R! J7 a  v# ~7 x  A" Z
while(postfix!=' '), b; J/ m3 C& r4 {8 E2 X
{
$ @: g- f- x6 K% d, j result = result*10+postfix[i++]-'0';) W4 k! t1 H/ ^4 T% z# ~
}
: i  g, a; z2 G6 [) i i++;& L  M6 u; |# _1 v
stack.push(result);
( M; j8 d* k% F/ o4 U" o }
% g6 I1 N0 q: a6 }6 @ else+ o* e  Y+ n3 N  [
{
4 S) l8 C' H5 j: w( \ if(postfix!=' ')
  ~, _% R& i- B! p  y5 \$ R {2 v% W8 [2 b& M3 b# K% t+ b! @
int y = stack.pop();4 y) p& F* R  N, M* l7 Z
int x = stack.pop();5 ?! w, X1 P$ s" W( i
switch(postfix)3 a* `$ l$ z# w5 t1 \0 V
{
) r+ F% D- k: D4 [ case'+':result = x + y;
* O! y( ?9 r+ I9 ?2 K- V; J break;
+ D, z" z( N% q. Q9 z case'-':result = x - y;" a" d& L1 `: D2 X7 g6 [6 F5 s
break;5 F. D  S9 g  e0 e
case'*':result = x *y;
3 X* p. s- S- Y, z! B break;1 k' d: q. q7 z+ d5 o
case'/':result = x / y;. }4 H1 m4 k" V. T
break;7 f% c$ [' [$ i
}8 J3 ^* r8 N7 B. p3 I) }$ G
stack.push(result);8 B  P  B( U& S
}1 ~/ H3 C' Z$ @/ Y" y8 Y" n
i++;8 E& \8 r5 x& \) `8 F3 z
}. _) u! d+ N+ N* F( _
}9 J5 e6 ]) C+ `
return stack.pop();
5 W4 o. Y) A' u9 w& r; r# R}' `+ K9 f& p3 T0 e1 H/ G1 P, P

9 `  i* B: H3 w$ T* qint main()
& o/ {( h5 z+ i# S% g9 Q5 z0 O0 N{3 X; D- ~. L* C4 B# D$ a4 |+ S: ~
//char *expstr = "121+10*(52-49+20)/((35-25)*2+10)";
  b& {+ a, ~* J1 d, h cout << "请输入表达式:";; m" B% F/ S7 }: O
//char *a ;7 T7 d/ N7 Q  X/ b& G8 \# Z
//cin >> *a;/ ~$ S4 q) F1 o! S
char expstr[20]={0};) E7 t; V5 ^$ K% Z9 ~
while(1)$ {( W' r2 ?4 O
{
* |8 ]5 S& I8 e& Z cin>>expstr;! Y/ O/ h( C! I8 n+ m. M3 ]6 J
char *postfix = toPostfix(expstr);
+ d" g: K$ r- e9 j2 q    cout << "expstr= "<<expstr << endl;
& w" x9 I! x) v( y$ Z3 ` cout << "postfix= "<<postfix<<endl;
, }8 Q1 y% u. ~0 J) t cout << "value= "<<value(postfix) << endl;7 m5 I" s4 Z. N2 t( w, {9 f
}; w+ N. R. |. S. {; C
return 0;6 \+ G' {( `( Z
}

外国人的数学素质是差啊

外国人的数学素质是差啊

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