P4711 「化学」相对分子质量

给你一个字符串让你输出相对分子质量（弱智字符串模拟）

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Code

#include<iostream>

#include<cstdio>

#include<queue>

#include<cstring>

#include<algorithm>

#include<climits>

#include<cstring>

#include<map>

typedef long long LL;

using namespace std;

int RD(){

    int out = 0,flag = 1;char c = getchar();

    while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}

    while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}

    return flag * out;

    }

const int maxn = 119;

map<string,double>M;

char s[maxn];

double ans,len;

double now;//

int main(){

    string x = "H";M[x] = 1;

    x = "C";M[x] = 12;

    x = "N";M[x] = 14;

    x = "O";M[x] = 16;

    x = "F";M[x] = 19;

    x = "Na";M[x] = 23;

    x = "Mg";M[x] = 24;

    x = "Al";M[x] = 27;

    x = "Si";M[x] = 28;

    x = "P";M[x] = 31;

    x = "S";M[x] = 32;

    x = "Cl";M[x] = 35.5;

    x = "K";M[x] = 39;

    x = "Ca";M[x] = 40;

    x = "Mn";M[x] = 55;

    x = "Fe";M[x] = 56;

    x = "Cu";M[x] = 64;

    x = "Zn";M[x] = 65;

    x = "Ag";M[x] = 108;

    x = "I";M[x] = 127;

    x = "Ba";M[x] = 137;

    x = "Hf";M[x] = 178.5;

    x = "Pt";M[x] = 195;

    x = "Au";M[x] = 197;

    x = "Hg";M[x] = 201;

    /////////

    cin>>s;len = strlen(s);

    for(int i = 0;i < len;i++){

        if(s[i] == '~'){//水合物

            i++;

            int numw = 0;

            while(s[i] <= '9' && s[i] >= '0'){numw = numw * 10 + s[i] - '0';i++;}

            ans += 18 * numw;

            if(numw == 0)ans += 18;

            break;

            }

        if(s[i] <= 'Z' && s[i] >= 'A'){//处理单个元素

            x = s[i];string temp;

            if(s[i + 1] >= 'a' && s[i + 1] <= 'z')temp = s[i + 1],i++;//是否为两个字母

            x = x + temp;//cout<<"x="<<x<<endl;

            for(map<string,double>::iterator it = M.begin(); it != M.end(); it++){

                string com = it->first;

                if(x == com){now = it->second;break;}

                }

            int tim = 0;

            if(s[i + 1] == '_'){

                while(s[i] != '{')i++;i++;//读取倍数

                while(s[i] != '}'){tim = tim * 10 + s[i] - '0';i++;}

                }

            if(tim)ans += tim * now;

            else ans += now;

            }

        if(s[i] == '('){

            i++;

            double ans1 = 0,now1 = 0;

            char n[maxn];int in = 0;

            while(s[i] != ')'){n[in++] = s[i++];}i++;

            for(int k = 0;k < in;k++){

                if(n[k] <= 'Z' && n[k] >= 'A'){

                    string a,b;a = n[k];

                    if(n[k + 1] <= 'z' && n[k + 1] >= 'a'){b = n[k + 1];k++;}

                    a = a + b;//cout<<a<<endl;

                    for(map<string,double>::iterator it = M.begin(); it != M.end(); it++){

                        string com = it->first;

                        if(com == a){now1 = it->second;break;}

                        }//printf("now1=%d\n",now1);

                    int tim1 = 0;

                    if(n[k + 1] == '_'){

                        while(n[k] != '{')k++;k++;//读取倍数

                        while(n[k] != '}'){tim1 = tim1 * 10 + n[k] - '0';k++;}

                        }

                    if(tim1)ans1 += tim1 * now1;

                    else ans1 += now1;

                    }

                }//printf("ans1=%d\n",ans1);

            if(s[i] == '_'){

                int tim2 = 0;

                while(s[i] != '{')i++;i++;

                while(s[i] != '}'){tim2 = tim2 * 10 + s[i] - '0';i++;}

                if(tim2)ans += ans1 * tim2;

                else ans += ans1;

                }

            }

        }

    if(ans == (int)ans)printf("%d\n",(int)ans);

    else printf("%.1lf",ans);

    return 0;

    }