sincerit-杭電oj 1237 簡單計算器
阿新 • • 發佈:2018-11-05
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26856 Accepted Submission(s): 9741
Problem Description
讀入一個只包含 +, -, *, / 的非負整數計算表示式,計算該表示式的值。
Input
測試輸入包含若干測試用例,每個測試用例佔一行,每行不超過200個字元,整數和運算子之間用一個空格分隔。沒有非法表示式。當一行中只有0時輸入結束,相應的結果不要輸出。
Output
對每個測試用例輸出1行,即該表示式的值,精確到小數點後2位。
Sample Input
1 + 2
4 + 2 * 5 - 7 / 11
0
Sample Output
3.00
13.36
#include <iostream>
#include <cstring>
#include <iomanip>
#include <stdio.h>
#include <stack>
using namespace std;
int main() {
double x;
char sign;
while (scanf("%lf", &x)) {
stack<double > num;
num.push(x);
char ch;
ch = getchar();
if (x == 0 && ch == '\n') break; // 特殊情況 0 + 1所以要加上ch=='\n'的判斷
scanf("%c", &sign);
getchar();
while (scanf("%lf", &x)) {
if (sign == '+') {
num.push(x);
} else if (sign == '-') {
num. push(-x);
} else if (sign == '*') {
double temp = num.top();
num.pop();
num.push(temp * x);
} else if (sign == '/') {
double t = num.top();
num.pop();
num.push(t*1.0 / x);
}
ch = getchar(); // 結束時數字後面有一個換行
if (ch == '\n') break;
scanf("%c", &sign);
getchar();
}
double sum = 0.0;
while (num.size()) {
sum += num.top();
num.pop();
}
//cout << fixed << setprecision(2) << sum << "\n";
printf("%.2f\n", sum);
}
return 0;
}
原先兩個棧模擬的不知道哪裡錯了
#include <iostream>
#include <stdio.h>
#include <stack>
#include <cstring>
#include <iomanip>
double cal(double x, double y, char ch) {
if (ch == '+') return y + x;
else if (ch == '-') return y - x;
else if (ch == '*') return y * x;
else return y / x;// 入棧再出來的資料剛好相反是y/x 不是x/y
}
int compare(char a, char b) {
if (a == '#') return 1;
else if (b == '*' || b == '/') {
if (a == '+' || a == '-') return 1;
}
return 0;
}
using namespace std;
int main() {
string str;
while (getline(cin, str)) {
if (str == "0") break;
stack<char> sign;
sign.push('#');
stack<double> num;
int k = 0;
for (int i = 0; i < str.size(); i++) {
if (str[i] >= '0' && str[i] <= '9') {
k = k * 10 + str[i] - '0';
} else if (str[i] == ' ') {
//cout << k << "\n";
num.push(k*1.0);
k = 0;
} else {
if (compare(sign.top(), str[i]) == 1) {
sign.push(str[i]);
} else {
double x = num.top();
num.pop();
double y = num.top();
num.pop();
num.push(cal(x, y, sign.top()));
sign.pop();
sign.push(str[i]); // 計算完了之後再把當前符號入棧
}
i++; // 符號後面的空格要跳過去
}
}
num.push(k*1.0); // 最後一個數據還在k中
while (sign.size() > 1) {
double a = num.top();
num.pop();
double b = num.top();
num.pop();
num.push(cal(a, b, sign.top()));
sign.pop();
}
//cout << fixed << setprecision(2) << num.top() << "\n";
printf("%.2f\n", num.top());
}
return 0;
}