int main()
{
    int m,n,t;
    m=1;
    printf("输入一个数计算其阶乘:\n");
    scanf("%d",&t);
    for(n=1;n<=t;n++)
        m=m*n;
    printf("阶乘：%d\n",m);
}

到34就计算为0

版主 帮我改一下好吗

数据类型不对
int main(void)
{
    float  m,t;
    int n;
    m=1;
    printf("输入一个数计算其阶乘:\n");
    scanf("%f",&t);
    for(n=1;n<=t;n++)
        m=m*n;
    printf("阶乘：%f\n",m);
return 0;
}


换成浮点型的 即可·

显然溢出，就是浮点数也有限制的

给你一个。并非原创的。

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

short mul(short a[],short d,short x)
{
    long i,y = 0; 

    for(i = 0;i < d; i++)
    {
        y += (long)a[i] * (long)x;
        a[i] = (short)(y % 10000);
        y /= 10000;
    } 

    a[d] = (short)y; 

    return d+!!y;   
} 

int main()
{
    time_t begin ,end;

    while(1)
    {

        time(&begin);

        long s;
        short *a,i,j,n,ws = 1;
   
        printf("N=");
        scanf("%d",&n);
   
        #define Pi 3.14159265358979323846L    //这个语句什么意思？
   
        s=(long)((log(2*Pi*n)/2+n*(log(n)-1))/log(10)+1);
   
        a=(short*)malloc((s/4+2)*sizeof(short));    //申请空间

        *a=1; 

        for(i=2;i<=n;i++)
            ws=mul(a,ws,i);        //函数调用了

        printf("%d!=%d",n,a[ws-1]); 

        for(j=ws-2;j>=0;j--)
            printf("%04d",a[j]);
   
        printf("\n");
   
        free(a);
   
        time(&end);
        printf("\nRunning time is %lf\n",difftime(end,begin));

        puts("continue(y/n)?");
        getchar();
        char sign = getchar();
        if('y' == sign || 'Y' == sign)
            continue;
        else
            break;
        }
    system("pause");

    return 0;

} 

樓主試試用double吧～

[ 本帖最后由 xx123321 于 2010-12-1 08:08 编辑 ]

用double的啊