UVA 1586 Molar Mass (c++)(字符串处理)(模拟)

题目大意就是给一个只含有C/H/O/N四个字母的分子式，求分子量。跟着题目意思来进行模拟就好了。重点与难点在于如何处理字母后一位数字以上的数字。写得略显繁杂。

#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
#define maxn 1000000+10
#include <ctype.h>
using namespace std ;

double mol[4] = {12.01,1.008,16.00,14.01} ;
/// C6H5OH
/// CO2
/// C12 H22 O11

int main(){
    int t ;
    std::cin >> t ;
    while ( t -- ){
        string str ;
        cin >> str ;
        int len = str.size() ;
        double num_C = 0 , num_H = 0 , num_O = 0 , num_N = 0 ;
        for ( int i = 0 ; i < len ; i ++ ){
            if ( i < len-1 ){
                if (isalpha(str[i])){
                    if ( str[i] == ‘C‘ && !isdigit(str[i+1]) ) num_C ++ ;
                    else if ( str[i] == ‘C‘ && isdigit(str[i+1]) ) {
                        double temp_C = 0 ;
                        int j = i+1 ;
                        while ( isdigit(str[j]) ){
                            temp_C = temp_C*10 + (str[j]-‘0‘) ;
                            j ++ ;
                        }
                        num_C += temp_C ;
                    }
                    if ( str[i] == ‘H‘ && !isdigit(str[i+1]) ) num_H ++ ;
                    else if ( str[i] == ‘H‘ && isdigit(str[i+1]) ) {
                        double temp_H = 0 ;
                        int j = i+1 ;
                        while ( isdigit(str[j]) ){
                            temp_H = temp_H*10 + (str[j]-‘0‘) ;
                            j ++ ;
                        }
                        num_H += temp_H ;
                    }
                    if ( str[i] == ‘O‘ && !isdigit(str[i+1]) ) num_O ++ ;
                    else if ( str[i] == ‘O‘ && isdigit(str[i+1]) ) {
                        double temp_O = 0 ;
                        int j = i+1 ;
                        while ( isdigit(str[j]) ){
                            temp_O = temp_O*10 + (str[j]-‘0‘) ;
                            j ++ ;
                        }
                        num_O += temp_O ;
                    }
                    if ( str[i] == ‘N‘ && !isdigit(str[i+1]) ) num_N ++ ;
                    else if ( str[i] == ‘N‘ && isdigit(str[i+1]) ) {
                        double temp_N = 0 ;
                        int j = i+1 ;
                        while ( isdigit(str[j]) ){
                            temp_N = temp_N*10 + (str[j]-‘0‘) ;
                            j ++ ;
                        }
                        num_N += temp_N ;
                    }
                }
            }
            else{
                if ( str[i] == ‘C‘ ) num_C ++ ;
                if ( str[i] == ‘H‘ ) num_H ++ ;
                if ( str[i] == ‘O‘ ) num_O ++ ;
                if ( str[i] == ‘N‘ ) num_N ++ ;
            }
        }
//        cout << num_C << " " << num_H << " " << num_O << " " << num_N << endl ;
        double ans = num_C*mol[0] + num_H*mol[1] + num_O*mol[2] + num_N*mol[3] ;
        printf("%.3lf\n" , ans) ;
    }
    return 0 ;
}

原文地址：https://www.cnblogs.com/Cantredo/p/9353652.html