2015
05-23

# Guess the Numbers

John has never been very good at maths. Due to his bad grades, his parents have sent him to the Academic Coalition of Mathematics (ACM). Despite the large amount of money his parents are spending on the ACM,
John does not pay much attention during classes. However, today he has begun to think about all the e ffort his parents are putting into his education, and he has started to feel somewhat. . . guilty. So he has made a decision: he is going to improve his maths grades!

However, no sooner had he resolved to pay attention than the lesson ended. So the only thing he has been able to do is to hurriedly copy the content of the blackboard in his notebook. Today, the teacher was explaining basic arithmetic expressions with unknowns. He vaguely remembers that his classmates have been substituting values into the unknowns to obtain the expressions’ results. However, in all the hurry, John has only written down expressions, values and results in a messy fashion. So he does not know which value comes with each unknown, or which result goes with each expression.
That is the reason he needs your help: he wants to know, given an expression, some values and a result, whether it is possible or not to assign those values to the unknowns in order for the expression to evaluate to the given result. The particular assignment of values does not matter to John, as he wants to do it by himself. He only wants to know whether it is possible or not.

Each test case in the input le consists of two lines:
 The fi rst line contains a sequence of natural numbers. The first one (1<=n<=5) is the number of unknowns that will occur in the expression. It is followed by a sequence of n integers v1 … vn (0<=vi<=50), which are the values to be assigned to the unknowns. Finally, there is an integer m (0<=m<=1000) representing the desired result of the evaluation of the expression.
 The second line contains an arithmetic expression composed of lowercase letters (a-z), brackets (( and )) and binary operators (+, -, *). This expression will contain n unknowns, represented by n di fferent lowercase letters, without repetitions. The expression will not contain any blanks and will always be syntactically correct, i.e. it is just an unknown or has the form (e1 op e2 ), where e1 and e2 are expressions and op is one of the three possible binary operators.
The input will finish with a dummy test case of just one line containing 0 0, which must not be processed.

Each test case in the input le consists of two lines:
 The fi rst line contains a sequence of natural numbers. The first one (1<=n<=5) is the number of unknowns that will occur in the expression. It is followed by a sequence of n integers v1 … vn (0<=vi<=50), which are the values to be assigned to the unknowns. Finally, there is an integer m (0<=m<=1000) representing the desired result of the evaluation of the expression.
 The second line contains an arithmetic expression composed of lowercase letters (a-z), brackets (( and )) and binary operators (+, -, *). This expression will contain n unknowns, represented by n di fferent lowercase letters, without repetitions. The expression will not contain any blanks and will always be syntactically correct, i.e. it is just an unknown or has the form (e1 op e2 ), where e1 and e2 are expressions and op is one of the three possible binary operators.
The input will finish with a dummy test case of just one line containing 0 0, which must not be processed.

3 2 3 4 14
((a+b)*c)
2 4 3 11
(a-b)
1 2 2
a
0 0

YES
NO
YES

1：准备2个栈，一个存符号（记为栈1），另一个存变量（记为栈2）；

2：依次读取中序表达式，

（1）如果当前字符是‘（’，直接进栈1；

（2）如果当前字符是‘）’，将符号栈栈顶元素依次弹入栈2，直至符号栈栈顶为‘（’，此时‘（’出栈；

（3）如果是变量，直接进栈2；

（4）如果是运算符

1》如果符号栈栈顶运算符优先级小于当前运算符优先级，或者符号栈为空或者符号栈栈顶元素为‘（’，直接进栈；

2》如果符号栈栈顶运算符优先级大于等于当前运算符优先级，将符号栈栈顶元素依次弹入栈2，直到符号栈栈顶运算符优先级小于当前运算符优先级，当前运算符进栈1；

#include <iostream>
#include<cstdio>
#include<cstring>

using namespace std;

char stack1[20],stack2[20],str[20];
int top1,top2;
int n,m;
int v[10];
int flag;
int fv[10];
int cur[10];

int cal()
{
int i,j;
int lcm[10],len = 0;
j = 0;
for(i = 0;i < top2;i ++)
{
if(islower(stack2[i]))
lcm[len ++] = cur[j ++];
else
{
len --;
switch(stack2[i])
{
case '+':lcm[len - 1] += lcm[len];
break;
case '-':lcm[len - 1] -= lcm[len];
break;
case '*':lcm[len - 1] *= lcm[len];
}
}
}
return lcm[0];
}

void dfs(int i,int cnt)
{
if(flag)
return;
if(cnt == n - 1)
{
if(cal() == m)
{
flag = 1;
}
return;
}
int j;
for(j = 0;j < n;j ++)
{
if(!fv[j])
{
fv[j] = 1;
cur[cnt + 1] = v[j];
dfs(j,cnt + 1);
fv[j] = 0;
}
}
}

int main()
{
int i;
char c;
while(scanf("%d",&n),n)
{
for(i = 0;i < n;i ++)
scanf("%d",&v[i]);
scanf("%d",&m);
top1 = top2 = 0;
scanf("%s",str);
for(i = 0;i < strlen(str);i ++)
{
switch(str[i])
{
case '(':stack1[top1 ++] = str[i];
break;
case ')':while(stack1[top1 - 1] != '(')
stack2[top2 ++] = stack1[--top1];
top1 --;
break;
case '*':if(top1 == 0 || stack1[top1 - 1] == '+'
|| stack1[top1 - 1] == '-' || stack1[top1 - 1] == '(')
stack1[top1 ++] = '*';
else
{
while(stack1[top1 - 1] == '*')
stack2[top2 ++] = stack1[--top1];
}
break;
case '+':
case '-':if(stack1[top1 - 1] == '(' || top1 == 0)
stack1[top1 ++] = str[i];
else
{
while(stack1[top1 - 1] == '*')//stack1[top1 - 1] == '+' || stack1[top1 - 1] == '-')//之前写的有问题，竟然能过。。。
stack2[top2 ++] = stack1[--top1];
}
break;
default:stack2[top2 ++] = str[i];
}
}
flag = 0;
for(i = 0;i < n;i ++)
{
memset(fv,0,sizeof(fv));
fv[i] = 1;
cur[0] = v[i];
dfs(i,0);
}
if(flag)
printf("YES\n");
else
printf("NO\n");
/*for(i = 0;i < top2;i ++)
printf("%c",stack2[i]);
printf("\n");*/
}
return 0;
}
//62MS	324K

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