#include "stdio.h"
#include "conio.h"
#define N  50
void main()
{
    int i, n;
    float x[N], y[N], sumx = 0, sumy = 0, equalx = 0, equaly = 0;
    double sumxy = 0, sumx_2 = 0, a, b;
    char answer;
    {
        printf("您一共要输入几组数据:");
        scanf("%d", &n);
        for (i = 1; i <= n; i++)
            {
                printf("请输入第%d组数据(x/y):", i);
                scanf("%f%f", &x[i], &y[i]);
            }
        for (i = 1; i <= n; i++)
            {
                sumx = sumx + x[i];
                sumy = sumy + y[i];
                sumxy = sumxy + x[i] * y[i];
                sumx_2 = sumx_2 + x[i] * x[i];
            }
        equalx = sumx / n;
        equaly = sumy / n;
        b = (sumxy - sumx * sumy / n) / (sumx_2 - sumx * sumx / n);
        a = equaly - b * equalx;
        printf("您所要求的线性方程为:y = %.3f - (%.3f) * x.\n", a, b);
        printf("您还想要继续计算下一组么(是请输入y):");
        /*answer = getche();*/
        scanf("%c", &answer);
    }while(answer == 'y');
    printf("\n按任意键退出！");
}
以上是我编的一个计算线性回归方程的程序，想让他能一次性进行多次计算，可怎么不等我输入y就自动退出循环了？各位高手帮我看一下哪里错了！！！！！

帮你运行了一下，没错！不过后面加上一句“getch();”或“system("pause");”才能实现你的“printf("\n按任意键退出！");”，当然，要是加上“system("pause");”就没必要写“printf("\n按任意键退出！");”了

用什么运行的，我的是visual C++ 6.0，怎么都运行不了啊！

还有，while循环前我刚好像少了一个do啊！麻烦解释一下！

还有，问题的关键是输入y后还能再输入一组数据进行计算啊，而不是计算一次就退出了！你确认有一次计算两次以上吗？

#include "stdio.h"
#include <string.h>
#define N  50
void main()
{
    int i, n;
    float x[N], y[N], sumx = 0, sumy = 0, equalx = 0, equaly = 0;
    double sumxy = 0, sumx_2 = 0, a, b;
    char answer;
    do
    {
        sumxy = 0, sumx_2 = 0,sumx = 0, sumy = 0, equalx = 0, equaly = 0;
        printf("您一共要输入几组数据:");
        scanf("%d", &n);
        for (i=1; i<=n;i++)
            {
                printf("请输入第%d组数据(x/y):",i);
                scanf("%f%f",&x[i],&y[i]);
            }
        for (i=1;i<=n;i++)
            {
                sumx=sumx+x[i];
                sumy=sumy+y[i];
                sumxy=sumxy+x[i]*y[i];
                sumx_2=sumx_2+x[i]*x[i];
            }
        equalx = sumx / n;
        equaly = sumy / n;
        b=(sumxy-sumx*sumy/ n)/(sumx_2-sumx*sumx/n);
        a=equaly-b*equalx;
        printf("您所要求的线性方程为:y = %.3f - (%.3f) * x.\n", a, b);
        printf("您还想要继续计算下一组么(是请输入y):");
        answer=getchar();
        scanf("%c",&answer);
    }
    while(answer=='y');
    printf("\n按任意键退出！");
}
虽然楼主已经结贴，但是我比较好事，哈哈哈，望大家不要见怪！
我修正了一下楼主的部分程序，希望对大家有所帮助吧。哈哈