UVA - 1586 Molar mass

//收获：运用isdigit()函数，解决了元素后面的原子个数可能有两位的情况。同时，每次遇到元素字母，计数工作都由函数完成，简化代码
//值得一提的是，这题后来又被我写了一次，充分利用了数组之间的一一对应，减少了许多代码...简洁许多啦！
//借鉴了下面blog的部分思路： http://blog.csdn.net/u014800748/article/details/32718177

#include <iostream>#include <cstring>#include <iomanip>#include <cctype>using namespace std;const int maxn = 200;char s[maxn];void getnumber(char*s, int i, int &a){if ( isdigit(s[i + 1]) && isdigit(s[i + 2]) )a += (s[i + 1] - '0') * 10 + (s[i + 2] - '0');elsea += (isdigit(s[i + 1]) ? (s[i + 1] - '0') : 1);}int main(){int t;cin >> t;while (t--){cin >> s;int len = strlen(s), a[4];double mass[4] = { 12.01, 1.008, 16.00, 14.01 }, sum = 0;char element[4] = { 'C', 'H', 'O', 'N'};memset(a, 0, sizeof (a));for (int i = 0; i < len; i++){// if (s[i] == 'C')// getnumber(s, i, a[0]);//// else if (s[i] == 'H')// getnumber(s, i, a[1]);//// else if (s[i] == 'O')// getnumber(s, i, a[2]);//// else if (s[i] == 'N')// getnumber(s, i, a[3]);for (int j = 0; j < 4; j++)if (s[i] == element[j]){getnumber(s, i, a[j]);continue;}}for (int i = 0; i < 4; i++)sum += (mass[i] * a[i]);cout << fixed << setprecision(3) << sum << endl;}return 0;}