UVA10268 498-bis多项式

Posted 2021-06-29 海岛Blog

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
$$a_{0}n{x}^{n-1}+a_1(n-1)x^{n-2}+...+a_{n-1}.$$
    All the input and output data will fit into integer, i.e. its absolute value will be less than 2³¹.
Input
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
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
7
1 -1
2
1 1 1
Sample Output
1
5

问题链接：UVA10268 498-bis
问题简述：（略）
问题分析：多项式问题，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA10268 498-bis */

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int x;
    while (scanf("%d", &x) == 1) {
        char c;
        getchar();
        c = getchar();

        int sum = 0, ans = 0, a;
        while (c != '\\n' && c != EOF) {
            if (c == '-' || isdigit(c)) {
                ungetc(c, stdin);
                scanf("%d", &a);
                ans = ans * x + sum;
                sum = sum * x + a;
            }
            c = getchar();
        }

        printf("%d\\n", ans);
    }

    return 0;
}