uva 327

题目大意：计算表达式的值

思路：刚开始有前缀后缀进行特殊处理，然后先将26个字母的值存放到一个数组中，如果出现前缀后缀需要进行++ – 就用一个flag进行++ – 最后的值加上flag就可以了。。 可是RE了。。不懂

#include <iostream>
using namespace std;
#include <cstring>
#include <deque>
#include <vector>
#include <algorithm>
char s[200];
int val[27];
int flag = 0;
vector<char> que;
deque<int> v;
void filter()
{
int i = 0;
for (int j = 0; j < strlen(s); j++)
{
if (s[j] != ' ')
s[i++] = s[j];
}
s[i] = '\0';
}
bool Prefix(int i) //前缀
{
if ((s[i - 1] == '+' && s[i - 2] == '+') || (s[i - 1] == '-' && s[i - 2] == '-'))
return true;

return false;
}
bool Suffix(int i) //后缀
{
if ((s[i + 1] == '+' && s[i + 2] == '+') || (s[i + 1] == '-' && s[i + 2] == '-'))
return true;
return false;
}
void solve()
{
int i = 0, j = 0,k=0,sum=0;
flag = 0;
if (!isalpha(s[0]))
{
if (s[0] == '+') //前缀
{
val[s[2] - 'a']++;
}
else if (s[0] == '-')
{
val[s[0] - 'a']--;
}
v.push_back(val[s[2] - 'a']); //这里写成s[0]
que.push_back(s[2]);
s[0] = s[1] = s[2] = ' ';
k = 2;
}
else
k = -1;
for (i = ++k; i < strlen(s); i++)
{
if (isalpha(s[i]))
{
if (i >= 2)
{
if (Prefix(i) && s[i - 1] == '+' )
{
val[s[i] - 'a']++;
s[i - 1] = s[i - 2] = ' ';
}
else if (Prefix(i) && s[i - 1] == '-')
{
val[s[i] - 'a']--;
s[i - 1] = s[i - 2] = ' ';
}
}
if (i + 2 < strlen(s))
{
if (Suffix(i) && s[i + 1] == '+' )
{
val[s[i] - 'a']++;
flag--;
s[i + 1] = s[i + 2] = ' ';
}
else if (Suffix(i) && s[i + 1] == '-')
{
val[s[i] - 'a']--;
flag++;
s[i + 1] = s[i + 2] = ' ';
}
}
v.push_back(val[s[i]-'a']);
que.push_back(s[i]);
s[i] = ' ';
}
}
filter();
for (i = 0; i < strlen(s); i++)
{
int a = v.front();
v.pop_front();
int b = v.front();
v.pop_front();
if (s[i] == '+')
sum = a + b;
else
sum = a - b;
v.push_front(sum);
}
printf("    value = %d\n",v.front()+flag);
sort(que.begin(), que.begin()+que.size());
for (i = 0; i < que.size(); i++)
printf("    %c = %d\n",que[i],val[que[i]-'a']);
que.clear();
v.clear();

}
int main()
{
int i = 0;
int j = 0;
int sum = 0;
while (gets(s))
{
for (i = 0; i < 26; i++)
val[i] = i + 1;
printf("Expression: %s\n", s);
filter();
solve();
}
}