UVaOJ 1586 - Molar mass

问题描述

解法一：模拟

扫描字符串，遍历到元素所代表的字符时，先看看后面是数字还是字符：如果后面直接是其它元素的字符，直接表明当前元素的原子数为 1；如果后面是一串数字，则求得该元素的原子个数，计算其质量并累加到分子质量 res 中。

#include <bits/stdc++.h>

using namespace std;

int main() {
  const map<char, float> TABLE = {
      {'C', 12.01}, {'H', 1.008}, {'O', 16.00}, {'N', 14.01}};

  int t;
  cin >> t;
  while (t--) {
    string formula;
    cin >> formula;

    float res = 0;
    for (int i = 0; i < formula.length(); ++i) {
      char atomic = formula[i];
      int cnt = 1;
      bool cnt_ommitted = true;
      while (i + 1 < formula.length() && isdigit(formula[i + 1])) {
        int digit = formula[++i] - '0';
        cnt = cnt_ommitted ? digit : 10 * cnt + digit;
        cnt_ommitted = false;
      }
      res += TABLE.at(atomic) * cnt;
    }
    cout << fixed << setprecision(3) << res << endl;
  }
  return 0;
}

#include <bits/stdc++.h>

using namespace std;

int main() {

const map<char, float> TABLE = {

{'C', 12.01}, {'H', 1.008}, {'O', 16.00}, {'N', 14.01}};

int t;

cin >> t;

while (t--) {

string formula;

cin >> formula;

float res = 0;

for (int i = 0; i < formula.length(); ++i) {

char atomic = formula[i];

int cnt = 1;

bool cnt_ommitted = true;

while (i + 1 < formula.length() && isdigit(formula[i + 1])) {

int digit = formula[++i] - '0';

cnt = cnt_ommitted ? digit : 10 * cnt + digit;

cnt_ommitted = false;

}

res += TABLE.at(atomic) * cnt;

}

cout << fixed << setprecision(3) << res << endl;

}

return 0;

}

Python 代码会 Runtime Error, 没有花时间去找原因：

TABLE = {'C': 12.01, 'H': 1.008, 'O': 16.00, 'N': 14.01}

t = int(input())
for _ in range(t):
    formula = input()
    res = pos = 0
    while pos < len(formula):
        atomic = formula[pos]
        cnt = 1
        cnt_omitted = True
        while (pos + 1 < len(formula) and formula[pos + 1].isdigit()):
            pos += 1
            digit = ord(formula[pos]) - ord('0')
            cnt = digit if cnt_omitted else 10 * cnt + digit
            cnt_omitted = False
        res += TABLE[atomic] * cnt
        pos += 1
    print(f"{res:.3f}")

TABLE = {'C': 12.01, 'H': 1.008, 'O': 16.00, 'N': 14.01}

t = int(input())

for _ in range(t):

formula = input()

res = pos = 0

while pos < len(formula):

atomic = formula[pos]

cnt = 1

cnt_omitted = True

while (pos + 1 < len(formula) and formula[pos + 1].isdigit()):

pos += 1

digit = ord(formula[pos]) - ord('0')

cnt = digit if cnt_omitted else 10 * cnt + digit

cnt_omitted = False

res += TABLE[atomic] * cnt

pos += 1

print(f"{res:.3f}")

AOAPC II UVaOJ 字符串