PAT基础编程题目-6-2 多项式求值

题目详情

【题目地址】：https://pintia.cn/problem-sets/14/problems/734

解答

C语言版

#include <stdio.h> #define MAXN 10 double f(int n, double a[], double x); int main() { int n, i; double a[MAXN], x; scanf("%d %lf", &n, &x); for (i = 0; i <= n; i++) scanf("%lf", &a[i]); printf("%.1f\n", f(n, a, x)); return 0; } double f(int n, double a[], double x) { double sum = 0; double y = 1; for (int i = 0; i <= n; i++) { for (int j = 0; j < i; j++) { y = x * y; break; } sum = sum + a[i] * y; } return sum; }

C++版

#include <iostream> #include <iomanip> using namespace std; #define MAXN 10 double f(int n, double a[], double x); int main() { int n; double a[MAXN], x; cin >> n >> x; for (int i = 0; i <= n; i++) { cin >> a[i]; } cout <<fixed<<setprecision(1)<<f(n, a, x); // 以固定浮点位显示，且以一位小数显示 return 0; } double f(int n, double a[], double x) { double sum = 0; double y = 1; for (int i = 0; i <= n; i++) { //for (int j = 0; j < i; j++) //{ // y = x * y; // break; // 每次多乘一个x就可以了 //} if (i > 0) y = x * y; sum = sum + a[i] * y; } return sum; }

Java版

import java.text.DecimalFormat; import java.util.Scanner; public class Main { private static final int MAXN = 10; private static double f(int n, double [] a, double x) { double sum = 0; double y = 1; for (int i = 0; i <= n; i++) { if( i > 0 ) y = x*y; sum = sum + a[i]*y; } return sum; } public static void main(String[] args) { int n = 0; double [] a = new double[MAXN]; //定义数组的方式 double x = 0; Scanner scanner = new Scanner(System.in); if(scanner.hasNext()) { n = scanner.nextInt(); x = scanner.nextDouble(); for (int i = 0; i <= n; i++) { a[i] = scanner.nextDouble(); } } scanner.close(); DecimalFormat decimalFormat = new DecimalFormat("#.0"); //保留小数点后一位 System.out.println(decimalFormat.format(f(n, a, x))); } }

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