# UVA 10268 498-bis
## 題目連結 [UVA 10268](https://vjudge.net/problem/UVA-10268)
### 題目內容
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
a~0~x^n^ + a~1~x^n−1^ + . . . + a~n−1~x + a~n~.
In this problem you should evaluate its derivative. Remember that derivative is defined as
a~0~nx^n−1^ + a~1~(n − 1)x^n−2^ + . . . + a~n−1~.
All the input and output data will fit into integer, i.e. its absolute value will be less than 2^31^.
### 輸入限制
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¿.
### 輸出限制
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.
### 解題思路
1.這裡有一個比較特別的for寫法，就是沒有終止條件，是因為不知道會輸入多少個數字，會利用getchar()=='\n'的方式來判斷是否輸入完畢，前提是輸入要用scanf(for迴圈裡面的n之所以不寫裡面是因為要計算輸入的數量)
2.函式裡的exp是把X帶進去後次方所得的數，所以每過一個迴圈就必須*X
3.是從倒數第二項開始算，因為最後一項微分後會消掉，而微分後的係數是max-i，拿總長度去減當下i值，可以得出未微分時的次方
### 程式碼

#include<bits/stdc++.h>
using namespace std;
int a[1000000];
int derivative(int x,int max){
    long long sum=0;
    int exp=1;
    for(int i=max-1;i>=0;i--){
        sum+=a[i]*exp*(max-i);
        exp*=x;
    }
    return sum;
}
int main(){
    int x;
    while(cin>>x){
        int n;
        for(n=0;;n++){
            scanf("%d",&a[n]);
            if(getchar()=='\n'){
                break;
            }
        }
        printf("%d\n",derivative(x,n));
    }
}

## 測資
### Sample input
7
1 -1
2
1 1 1
### Sample output
1
5
## 中文題目連結 [zerojudge f444](https://zerojudge.tw/ShowProblem?problemid=f444)