高中生程式解題系統 f444: 10268 - 498-bis－資工趴趴熊的小天地

Content

給你一組方程式和X求微分後的值

Looking throw the “Online Judge’s Problem Set Archive” I found a very interesting problem number 498, titled “Polly the Polynomial”. Frankly speaking, I did not solve it, but I derived from it this problem.

Everything in this problem is a derivative of something from 498. In particular, 498 was “... designed to help you remember ... basic algebra skills, make the world a better place, etc., etc.”. This problem is designed to help you remember basic derivation algebra skills, increase the speed in which world becomes a better place, etc., etc.

In 498 you had to evaluate the values of polynomial

a0x^n + a1x^n−1 + . . . + an−1x + an.

In this problem you should evaluate its derivative. Remember that derivative is defined as

a0nx^n−1 + a1(n − 1)x^n−2 + . . . + an−1.

All the input and output data will fit into integer, i.e. its absolute value will be less than 2^31.

Input

每組測資有兩行第一行為X 第二行為方程式(降冪)

第二行最後面會先有空格在換行

Your program should accept an even number of lines of text. Each pair of lines will represent one
problem. The first line will contain one integer - a value for x. The second line will contain a list of
integers a0, a1, ..., an−1, an, which represent a set of polynomial coefficients.
Input is terminated by EOF.

Output

求出微分後的值

For each pair of lines, your program should evaluate the derivative of polynomial for the given value x
and output it in a single line.

Sample Input #1

Sample Output #1

1
5

python:

while True:
    try:
        x = int(input()) #輸入的X值
        A = list(map(int, input().strip().split()))  #這裡放置 1X^2 + 1X^1 + 1X^0  的係數
        ans = 0
        n = len(A) - 1  #n=2
        for i in range(n):
            ans += A[i] * (n-i) * pow(x, n-1-i)  
        #第一個來說 係數 * 微分下來的次方 *新次方(要-1) 
        print(int(ans))
    except:
        break

C++

#include <iostream>
#include <sstream>
#include <vector>
#include <algorithm>
#include <string>
using namespace std;
 
int main() {
    int n;
	string s;
	int g;
	vector<int> v;
    while (cin >> n ){
		getline(cin,s);
		getline(cin,s);
		//cout<<"s "<<s<<'\n';
		stringstream ss(s);
		v.clear();
		while (ss>>g)
		{
			v.push_back(g);
		}
		
		v.pop_back();
		reverse(v.begin(),v.end());// -1 1
		long long int mul=1; 
		int ans=0;
		for (int i = 0; i < v.size(); i++)
		{

			//cout << "i: " << i << " " << v[i] << " " << ans << " " << mul << "\n";
			ans += v[i]*(i+1)*mul;
            mul *= n;
            //cout << "i: " << i << " " << v[i] << " " << ans << " " << mul << "\n";
		}
		cout<<ans<<'\n'	;
	}
    return 0;
}
/*
7
1 -1 
2
1 1 1 
*/