[刷题] 多项式计算

1、写一个函数PrintN，使得传入一个N，打印从1到N的全部整数

#include<stdio.h>
//循环实现
void PrintN(int N){
    int i;
    for(i = 1;i<=N;i++){
        printf("%d
",i);
    }
    return;
}

//递归实现
void PrintN1(int N){
    if(N){
        PrintN1(N-1);
        printf("%d
",N);
    }
    return;
}

int main(){
    int N;
    scanf("%d",&N); 
    PrintN(N);
    return 0;
}

2、写程序计算给定多项式在给定点x处的值f(x)=a0+a1*x+...+an-1*x^(n-1)+an*x^n

将多项式写成如下形式：f(x)=a0+x(a1+x(...(an-1+x(an))...))

#include <stdio.h>
#include <time.h>
#include <math.h>

clock_t start,stop;
double duration;
#define MAXN 10    /*多项式次数*/ 
#define MAXK 1e7    /*被测函数最大重复调用次数*/ 
double f1(int n,double a[],double x);
double f2(int n,double a[],double x);

int main(){
    int i;
    double a[MAXN];
    for (i=0;i<MAXN;i++) a[i]=(double)i;
    /*测试f1*/
    start = clock();    /*开始计时*/ 
    for(i=0;i<MAXK;i++)
        f1(MAXN-1,a,1.1);        /*被测试函数*/
    stop = clock();        /*停止计时*/
    duration = ((double)(stop-start))/CLK_TCK/MAXK;
    printf("ticks1 = %f
",(double)(stop-start));
    printf("duration1 = %6.2e
",duration); 
    /*测试f2*/
    start = clock();
    for(i=0;i<MAXK;i++)
        f2(MAXN-1,a,1.1);    
    stop = clock();
    duration = ((double)(stop-start))/CLK_TCK/MAXK;
    printf("ticks2 = %f
",(double)(stop-start));
    printf("duration2 = %6.2e
",duration); 
    
    return 0;
}

double f1(int n,double a[],double x){
    int i;
    double p = a[0];
    for(i = 1;i<=n;i++)
    p+=(a[i]*pow(x,i));
    return p;
}

double f2(int n,double a[],double x){
    int i;
    double p = a[n];
    for(i = 1;i>0;i--)
    p=a[i-1]+x*p;
    return p;
}

 结果：