化学方程式

复制import re
from collections import Counter

# 解析化学式，返回元素计数
# 例如 'H2O' → {'H': 2, 'O': 1}
def parse_formula(formula: str) -> dict:
    pattern = r'([A-Z][a-z]*)(\d*)'
    counts = Counter()
    for elem, num in re.findall(pattern, formula):
        counts[elem] += int(num) if num else 1
    return counts

# 示例：配平 H2 + O2 → H2O
# 这里用简单线性代数求解系数（仅演示，实际需矩阵求解）
reactants = ['H2', 'O2']
products = ['H2O']

# 手动配平结果：2H2 + O2 → 2H2O
coeffs = {'H2': 2, 'O2': 1, 'H2O': 2}
print('配平方程:')
print(f"{coeffs['H2']}{reactants[0]} + {coeffs['O2']}{reactants[1]} → {coeffs['H2O']}{products[0]}")
# 输出: 2H2 + O2 → 2H2O

复制package main

import (
	"fmt"
	"regexp"
	"strconv"
)

// 解析化学式，返回元素计数
func parseFormula(formula string) map[string]int {
	re := regexp.MustCompile(`([A-Z][a-z]*)(\d*)`)
	matches := re.FindAllStringSubmatch(formula, -1)
	counts := make(map[string]int)
	for _, m := range matches {
		elem := m[1]
		num := 1
		if m[2] != "" {
			num, _ = strconv.Atoi(m[2])
		}
		counts[elem] += num
	}
	return counts
}

func main() {
	// 示例：解析 H2O
	result := parseFormula("H2O")
	fmt.Println(result) // map[H:2 O:1]

	// 配平示例：2H2 + O2 → 2H2O
	fmt.Println("配平方程: 2H2 + O2 → 2H2O")
}

复制// 解析化学式，返回元素计数
function parseFormula(formula) {
  const pattern = /([A-Z][a-z]*)(\d*)/g;
  const counts = {};
  let match;
  while ((match = pattern.exec(formula)) !== null) {
    const elem = match[1];
    const num = match[2] ? parseInt(match[2], 10) : 1;
    counts[elem] = (counts[elem] || 0) + num;
  }
  return counts;
}

// 示例
console.log(parseFormula('H2O')); // { H: 2, O: 1 }
console.log(parseFormula('C6H12O6')); // { C: 6, H: 12, O: 6 }

// 配平方程示例（手动系数）
const balanced = '2H2 + O2 → 2H2O';
console.log('配平方程:', balanced);