多项式求和

$$\sum = i*x^i$$

#include <stdio.h>
#include <time.h>
#include <math.h>
#define MAXN 10
#define MAXK 1e7
clock_t start, stop;
double duration;

double f1(int n, double a[], double x);//直接的方法
double f2(int n, double a[], double x);//从里面到外面的方法

int main()
{
    int i;
    double a[MAXN], sum1, sum2;
    for(i=0; i<MAXN; i++)
        a[i] = (double)i;

    start = clock();
    for(i=0; i<MAXK; i++)
        sum1=f1(MAXN-1, a, 1.1);
    stop = clock();
    duration = ((double)(stop-start))/CLK_TCK/MAXK;
    printf("sum1 = %f\n", sum1);
    printf("ticks1 = %f\n",(double)(stop-start));
    printf("duration = %6.2e\n\n\n", duration);

    start = clock();
    for(i=0; i<MAXK; i++)
        sum2=f2(MAXN-1, a, 1.1);
    stop = clock();
    duration = ((double)(stop-start))/CLK_TCK/MAXK;
    printf("sum2 = %f\n", sum2);
    printf("ticks2 = %f\n",(double)(stop-start));
    printf("duration = %6.2e\n", duration);

    return 0;
}

double f1(int n, double a[], double x)
//算法复杂度( 1+2+3...+n=(n^2+n)/2) 其实就是算乘法的次数，Tn=C1n^2+C2n   
{
    double sum=0;
    int i;
    for(i=0; i<=n; i++)
    {
        sum += (a[i]*pow(x,i));//这里+=后面加个大括号更保险
    }
    return sum;

}
double f2(int n, double a[], double x)
//算法复杂度n Tn=C*n
{
    double sum=a[n];
    int i;
    for(i=n; i>0; i--)//注意这里是>0
    {
        sum = a[i-1] + sum*x;//这里也巧妙
    }
    return sum;
}