UVa 1586 Molar Mass
題意:
輸入一個分子式算出分子量題解:
輸入的分子式含有C、O、N、H四個元素
係數不超過99
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| //UVa 1586 - Molar Mass | |
| #include <cstdio> | |
| #include <cstring> | |
| #include <cctype> //用到isAlpha(), isDigit() | |
| const double C = 12.010; | |
| const double H = 1.008; | |
| const double N = 14.010; | |
| const double O = 16.000; | |
| void add(char emt, int mole, double& mass) { | |
| switch (emt) { | |
| case 'C' : | |
| mass += C*mole; | |
| break; | |
| case 'H': | |
| mass += H*mole; | |
| break; | |
| case 'N': | |
| mass += N*mole; | |
| break; | |
| case 'O': | |
| mass += O*mole; | |
| break; | |
| default: | |
| break; | |
| } | |
| } | |
| int main() { | |
| int t; | |
| scanf("%d", &t); | |
| for (int i = 0; i < t; ++i) { | |
| char molar[85]; | |
| double mass = 0; | |
| scanf("%s", molar); | |
| int len = strlen(molar); | |
| for (int j = 0; j < len; ++j) { | |
| char emt = molar[j]; | |
| char p1 = molar[j+1], p2 = molar[j+2]; //係數十位與個位 | |
| if(isalpha(emt)) { //先判斷是否為字母，false表示是係數 | |
| if(isdigit(p1)) { //判斷是否有係數 | |
| if(isdigit(p2)) { //再判斷是否為雙位係數 | |
| add(emt, (p1 - '0')*10 + (p2 - '0'), mass); | |
| j = j+2; //增加效率(可省略) | |
| } else { | |
| add(emt, p1 - '0', mass); | |
| j++; | |
| } | |
| } else { //沒係數表示係數是1 | |
| add(emt, 1, mass); | |
| } | |
| } | |
| } | |
| printf("%.3f\n", mass); | |
| } | |
| return 0; | |
| } |
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