本题要求实现一个函数,计算阶数为n,系数为a[0] ... a[n]的多项式f(x)=∑i=0n(a[i]×xi) 在x点的值. 函数接口定义: double f( int n, double a[], double x ); 其中n是多项式的阶数,a[]中存储系数,x是给定点.函数须返回多项式f(x)的值. 裁判测试程序样例: #include <stdio.h> #define MAXN 10 double f( int n, double a[], double x );
题目地址 :多项式求和 /* #include<stdio.h> int main() { int n,b; double a[110],x; double z; int i,j; int f; for (i=1;i<101;i++) a[i]=(double)1/(double)i; scanf("%d",&n); while(n--) { scanf("%d",&b); z = 0; f = 1; //printf("
Description Given any integer 0 <= n <= 10000 not divisible by 2 or 5, some multiple of n is a number which in decimal notation is a sequence of 1's. How many digits are in the smallest such a multiple of n? Input Each line contains a number n.