ACM模版
描述
题解
这是第八届河南 ACM 省赛的第一题好像，和去年的第一题表达式求值有相似之处，但是由于这个题目明确提示，不会有嵌套，所以不必使用栈，直接模拟一遍就可以了。感觉比去年的那个要简单很多。可是难度等级却比去年的还高……
代码
#include <iostream>
#include <cstring>
#include <map>
#include <cstdio>
#include <string>
using namespace std;
map<string, int> msi;
void init()
{
msi[
"N"] =
14;
msi[
"C"] =
12;
msi[
"O"] =
16;
msi[
"Cl"] =
35;
msi[
"S"] =
32;
msi[
"H"] =
2;
msi[
"Al"] =
27;
msi[
"Ca"] =
40;
msi[
"Zn"] =
65;
msi[
"Na"] =
23;
}
int main()
{
init();
int K;
cin >> K;
string s;
while (K--)
{
cin >> s;
int i =
0;
while (s[i] !=
'=')
{
i++;
}
i++;
int coe =
0;
while (s[i] >=
'0' && s[i] <=
'9')
{
coe *=
10;
coe += s[i] -
'0';
i++;
}
if (coe ==
0)
{
coe =
1;
}
int res =
0;
int tmp, n;
while (s[i] !=
'+' && i < s.length())
{
if (s[i] ==
'(')
{
i++;
tmp =
0;
int tmp_;
while (s[i] !=
')')
{
if (s[i] ==
'N')
{
if (s[i +
1] ==
'a')
{
tmp_ = msi[
"Na"];
i++;
}
else
{
tmp_ = msi[
"N"];
}
}
else if (s[i] ==
'C')
{
if (s[i +
1] ==
'a')
{
tmp_ = msi[
"Ca"];
i++;
}
else if (s[i +
1] ==
'l')
{
tmp_ = msi[
"Cl"];
i++;
}
else
{
tmp_ = msi[
"C"];
}
}
else if (s[i] ==
'A')
{
tmp_ = msi[
"Al"];
i++;
}
else if (s[i] ==
'Z')
{
tmp_ = msi[
"Zn"];
i++;
}
else
{
string a =
"";
tmp_ = msi[a + s[i]];
}
i++;
int n_ =
0;
while (s[i] >=
'0' && s[i] <=
'9')
{
n_ *=
10;
n_ += s[i] -
'0';
i++;
}
if (n_ ==
0)
{
n_ =
1;
}
tmp += tmp_ * n_;
}
goto tag;
}
if (s[i] ==
'N')
{
if (s[i +
1] ==
'a')
{
tmp = msi[
"Na"];
i++;
}
else
{
tmp = msi[
"N"];
}
}
else if (s[i] ==
'C')
{
if (s[i +
1] ==
'a')
{
tmp = msi[
"Ca"];
i++;
}
else if (s[i +
1] ==
'l')
{
tmp = msi[
"Cl"];
i++;
}
else
{
tmp = msi[
"C"];
}
}
else if (s[i] ==
'A')
{
tmp = msi[
"Al"];
i++;
}
else if (s[i] ==
'Z')
{
tmp = msi[
"Zn"];
i++;
}
else
{
string a =
"";
tmp = msi[a + s[i]];
}
tag:
i++;
n =
0;
while (s[i] >=
'0' && s[i] <=
'9')
{
n *=
10;
n += s[i] -
'0';
i++;
}
if (n ==
0)
{
n =
1;
}
res += tmp * n;
}
printf(
"d\n", res * coe);
}
return 0;
}
