第1个回答 2010-05-31
#include<stdio.h>
#include<stdlib.h>
int main()
{
int n = 6;
double aaverage=0,paverage=0,psquare=0,pmultiplya=0;
double a[]={181,197,235,270,283,292};
double p[]={36.9,46.7,63.7,77.8,84,87.5};
for(int i=0;i<n;i++)
{
aaverage+=a[i];
paverage+=p[i];
psquare+=p[i]*p[i];
pmultiplya+=a[i]*p[i];
}
aaverage/=n;
paverage/=n;
double aa = (pmultiplya-n*aaverage*paverage)/(psquare-n*paverage*paverage);
double b = aaverage-aa*paverage;
printf("a=%lf,b=%lf",aa,b);
}本回答被提问者采纳