nyoj 挑战密室 模拟题

题目意思为求化学反应方程式等号后面一个物质的原子量，主要注意处理括号问题。

这里有个全局变量 j 要注意一下

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn=1e3+10;
const int INF=0x3f3f3f3f;
typedef long long ll;
char s[100];
int j;
int judge()
{
    int res;
    if(s[j]=='N')
    {
        if(s[j+1]=='a')
        {
            res=23;
            j++;
        }
        else
            res=14;
    }
    else if(s[j]=='C')
    {
        if(s[j+1]=='l')
        {
            j++;
            res=35;
        }
        else if(s[j+1]=='a')
        {
            j++;
            res=40;
        }
        else
            res=12;
    }
    else if(s[j]=='O')
        res=16;
    else if(s[j]=='S')
        res=32;
    else if(s[j]=='H')
        res=2;
    else if(s[j]=='A'&&s[j+1]=='l')
    {
        res=27;
        j++;
    }
    else if(s[j]=='Z'&&s[j+1]=='n')
    {
        res=65;
        j++;
    }
    j++;
    int bs=1;
    if(s[j]>='0'&&s[j]<='9')
    {
        bs=0;
        while(s[j]>='0'&&s[j]<='9')
        {
            bs=bs*10+(s[j]-'0');
            j++;
        }
    }
    return res*bs;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",s);
        int len=strlen(s);
        int f=0;
        for(int i=0; iif(s[i]=='=')
            {
                f=i+1;
                break;
            }
        }
        int ff=1;
        j=f;
        if(s[j]>='0'&&s[j]<='9')
        {
            ff=0;
            while(s[j]>='0'&&s[j]<='9')
            {
                ff=ff*10+(s[j]-'0');
                j++;
            }
        }
        //printf("ff=%d\n",ff);
        int sum=0,ys,kh,bs;
        while(s[j]!='+'&&jif(s[j]>='A'&&s[j]<='Z')
            {
                sum+=judge();
            }
            else if(s[j]=='(')
            {
                //puts("______\n");
                int kh=0;
                j++;
                while(s[j]>='A'&&s[j]<='Z')
                {
                    kh+=judge();
                }
                if(s[j]==')')
                {
                    j++;
                     bs=1;
                    if(s[j]>='0'&&s[j]<='9')
                    {
                        bs=0;
                        while(s[j]>='0'&&s[j]<='9')
                        {
                            bs=bs*10+(s[j]-'0');
                            j++;
                        }
                    }
                }
                //printf("()=====%d %d\n",kh,bs);
                sum+=kh*bs;
            }
        }
        sum*=ff;
        printf("%04d\n",sum);
    }
}