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紫书上习题3-2，这道题我用的普通的 for 循环做的，逐个判断字符，若该字符为 C , H , O , N ,则判断该字符的后两位是否为数字，是数字就乘以该字符的原子量，最后求和；

v#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <iomanip>#include <algorithm>using namespace std;int main(){    char C,H,O,N;    string S;int n,p;    while(cin>>n){      while(n--){        cin>>S;        int len=S.length();double sum=0;        for(int i=0;i<len;i++){            if(S[i]=='C'){                if(S[i+1]>=49&&S[i+1]<=57){                    if(S[i+2]>=48&&S[i+2]<=57)                        p=10*(S[i+1]-'0')+(S[i+2]-'0');                    else                        p=S[i+1]-'0';                }                else                    p=1;                sum=sum+12.010*p;            }            else if(S[i]=='H'){                if(S[i+1]>=49&&S[i+1]<=57){                    if(S[i+2]>=48&&S[i+2]<=57)                        p=10*(S[i+1]-'0')+(S[i+2]-'0');                    else                        p=S[i+1]-'0';                }                else                    p=1;                sum=sum+1.008*p;            }            else if(S[i]=='O'){                if(S[i+1]>=49&&S[i+1]<=57){                    if(S[i+2]>=48&&S[i+2]<=57)                        p=10*(S[i+1]-'0')+(S[i+2]-'0');                    else                        p=S[i+1]-48;                }                else                    p=1;                sum=sum+16.000*p;            }            else if(S[i]=='N'){                if(S[i+1]>=49&&S[i+1]<=57){                    if(S[i+2]>=48&&S[i+2]<=57)                        p=10*(S[i+1]-'0')+(S[i+2]-'0');                    else                        p=S[i+1]-'0';                }                else                    p=1;                sum=sum+14.01*p;            }            else                continue;        }        cout<<fixed<<setprecision(3)<<sum<<endl;      }    }    return 0;}

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