#include "stdio.h"
#include "conio.h"

void input(void)
{
        int a, i ;
        float sum = 0.0,i1=0.0;
        float score[100];

        printf("input ren shu:");    /*输入人数*/
        scanf("%d", &a);
        for (i = 0; i < a; i++) {
                printf("input %d de chengji:  ", i+1);
                scanf("%f", &score[i]);
        }
        for (i = 0; i < a; i++)
                sum += score[i];
        printf("sum is :%2f\n", sum);    /*求总分*/
        printf("\n");
        printf("Aver is :%f\n",sum/a);  /*求平均分*/
        printf("\n");
        for (i =0;i < a;  i++) {        /*求及格人数和及格率*/
                if (score[i]>=60)
                i1++;
       }
        printf("jige renshu is %d ren.\n",i1);
        printf("\n");
        printf("jige renshu de is %f%.\n",i1/a);  /*这里不知道怎么修改了..得不到百分数的值,最好是2位小数.*/
}

int main(void)
{
        input();
        getchar();
        while(1) {getchar();}     
}

i1 到底是定义成什么类型.
定为float则不能够求出及格人数.

我知道了..用强制转换.
printf("jige renshu de is %f%.\n",(float)i1/a(float));

结果是6位小数的,有没有办法四舍五入保留2位小数的??

谢谢,可以了用了..
为什么*100后就能四舍五入了呢?*1000是保留三位小数吧.