寻找给定多项式的不定积分的程序

给定一个多项式字符串str，任务是给定的字符串整合，整合之后打印字符串。

注意：输入格式是这样的，在术语和“+”符号之间有一个空格。

例子：

Input: str = “4X³ + 3X¹ + 2X²”
Output: X⁴ + (3/2)X² + (2/3)X³ + C

Input: str = “5X³ + 7X¹ + 2X² + 1X⁰”
Output: (5/4)X⁴ + (7/2)X² + (2/3)X³ + X^q + C

方法：思想是观察当给定的方程由多个多项式组成时$p(x) = p1(x) + p2(x)$ , 给定多项式的积分$P(x) = P1(x) + P2(x)$ .

还知道不定积分$p(x) = AX^N$是$P(x) = \frac{A*X^{N + 1}}{N + 1 }+ C$ .

因此，我们拆分给定的字符串并整合其中的每个术语。

下面是上述方法的实现：

// C++ program to find the indefinite
// integral of the given polynomial
  
#include "bits/stdc++.h"
#define MOD (1e9 + 7);
using ll = int64_t;
using ull = uint64_t;
#define ll long long
using namespace std;
  
// Function to perform the integral
// of each term
string inteTerm(string pTerm)
{
    // Get the coefficient
    string coeffStr = "", S = "";
    int i;
  
    // Loop to iterate and get the
    // Coefficient
    for (i = 0; pTerm[i] != 'x'; i++)
        coeffStr.push_back(pTerm[i]);
    long long coeff
        = atol(coeffStr.c_str());
  
    string powStr = "";
  
    // Loop to find the power
    // of the term
    for (i = i + 2; i != pTerm.size(); i++)
        powStr.push_back(pTerm[i]);
  
    long long power
        = atol(powStr.c_str());
    string a, b;
    ostringstream str1, str2;
  
    // For ax^n, we find a*x^(n+1)/(n+1)
    str1 << coeff;
    a = str1.str();
    power++;
    str2 << power;
    b = str2.str();
    S += "(" + a + "/" + b + ")X^" + b;
  
    return S;
}
  
// Function to find the indefinite
// integral of the given polynomial
string integrationVal(string& poly)
{
  
    // We use istringstream to get the
    // input in tokens
    istringstream is(poly);
  
    string pTerm, S = "";
  
    // Loop to iterate through
    // every term
    while (is >> pTerm) {
  
        // If the token = '+' then
        // continue with the string
        if (pTerm == "+") {
            S += " + ";
            continue;
        }
  
        if (pTerm == "-") {
            S += " - ";
            continue;
        }
  
        // Otherwise find
        // the integration of
        // that particular term
        else
            S += inteTerm(pTerm);
    }
    return S;
}
  
// Driver code
int main()
{
    string str
        = "5x^3 + 7x^1 + 2x^2 + 1x^0";
    cout << integrationVal(str)
         << " + C ";
    return 0;
}

输出：

(5/4)X^4 + (7/2)X^2 + (2/3)X^3 + (1/1)X^1 + C