多项式导数的C ++程序
给定一个包含多项式项的字符串，任务是评估该多项式的导数。
什么是多项式？
多项式有两个词：-“Poly”表示“许多”，“nomial”表示“项”，其中包括许多项。多项式表达式是包含变量，系数和指数的表达式，它仅涉及变量的加，乘和减运算。
多项式示例
x2+x+1
多项式p(x)=mx^n的导数为-
m*n*x^(n-1)
示例
Input: str = "2x^3 +1x^1 + 3x^2" val = 2 Output: 37 Explanation: 6x^2 + 1x^0 + 6x^1 Putting x = 2 6*4 + 1 + 6*2 = 24 + 1 + 12 = 37 Input: str = “1x^3” val = 2 Output: 12 Explanation: 1 * 3 *x^2 Putting x = 2 3 * 4 = 12
我们将用来解决上述问题的方法-
将输入作为字符串和值x
现在遍历字符串并检查数字和变量。
继续添加和遍历字符串，直到找到“+”。
然后m*n*x^(n-1)。
返回结果。
算法
Start
Step 1-> In function long long term(string polyterm, long long val)
Declare and initialize coeffStr = "”
Declare i
Loop For i = 0 and polyterm[i] != 'x' and i++
Call coeffStr.push_back(polyterm[i])
Set coeff = atol(coeffStr.c_str()
Declare and initialize powStr = ""
Loop For i = i + 2 and i != polyterm.size() and i++ powStr.push_back(polyterm[i])
Set power = atol(powStr.c_str());
Return coeff * power * pow(val, power - 1)
Step 2-> In function long long value(string& str, int val)
Set ans = 0
Call istringstream is(str)
Declare string polyterm
Loop While is >> polyterm
If polyterm == "+” then,
Continue
Else
Set ans = (ans + term(polyterm, val))
Return ans
Step 3-> In function int main() Declare and initialize str = "2x^3 + 1x^1 + 3x^2"
Declare and initialize val = 2
Print the value received by value(str, val)
Stop示例
#include
using namespace std;
long long term(string polyterm, long long val) {
//找到系数
string coeffStr = "";
int i;
for (i = 0; polyterm[i] != 'x'; i++)
coeffStr.push_back(polyterm[i]);
long long coeff = atol(coeffStr.c_str());
//获得幂值
string powStr = "";
for (i = i + 2; i != polyterm.size(); i++)
powStr.push_back(polyterm[i]);
long long power = atol(powStr.c_str());
//对于ax^n，我们返回a(n-1)x^(n-1)
return coeff * power * pow(val, power - 1);
}
long long value(string& str, int val) {
long long ans = 0;
//使用istringstream获取令牌中的输入
istringstream is(str);
string polyterm;
while (is >> polyterm) {
//检查令牌是否等于“+”，然后
//继续字符串
if (polyterm == "+")
continue;
//否则找到那个的导数
//特定术语
else
ans = (ans + term(polyterm, val));
}
return ans;
}
//主要功能
int main() {
string str = "2x^3 + 1x^1 + 3x^2";
int val = 2;
cout << value(str, val);
return 0;
}输出结果
37