程序的目的是实现：a+aa+aaa+... ...+aaa..aaa(n个a)的和，如n=3，a=2,则为：2+22+222。我的代码如下：
#include <math.h>
main ()
{
    long int a, temp=0, add=0, sum=0;
    int n, i;
    printf("Please input positive integers 'a' and 'n':\n");
    scanf("%ld %d", &a, &n);
    while (a<=0 || n<=0)
    {
        printf("Sorry! 'a' or 'n' doesn't fit. Please input 'a' & 'n' again:\n");
        scanf("%ld %d", &a, &n);
    }
    while (a>=10)
    {
        printf("Sorry! 'a' can not be larger than 9. Please input 'a' & 'n' again:\n");
        scanf("%ld %d", &a, &n);
    }
    for (i=n; i>0; i--)
    {
        temp=temp+a*pow(10,i-1);
    }
    add=(int)(temp);
    sum=add;
    while (add>9)
    {
        add=(int)(add/10);
        sum=sum+add;
    }
    printf("%d+%d%d+...+%d=%d.\n",a,a,a,(int)(temp),sum);
}
编译通过，测试：a=2, n=6，结果完全不对，a甚至还等于0了。我郁闷了，到底错哪了？怎么也调不出来！谢谢了！

#include <math.h>
main ()
{
    long int a, temp=0, add=0, sum=0;
    int n, i;
    printf("Please input positive integers 'a' and 'n':\n");
    scanf("%ld %d", &a, &n);
    while (a<=0 || n<=0)
    {
        printf("Sorry! 'a' or 'n' doesn't fit. Please input 'a' & 'n' again:\n");
        scanf("%ld %d", &a, &n);
    }
    while (a>=10)
    {
        printf("Sorry! 'a' can not be larger than 9. Please input 'a' & 'n' again:\n");
        scanf("%ld %d", &a, &n);
    }
    for (i=n; i>0; i--)
    {
        temp=temp+a*pow(10,i-1);//将double转换成int时发生了错误，如30000000.00000转换成了29999999
    }
    add=(int)(temp);
    sum=add;
    while (add>9)
    {
        add=(int)(add/10);
        sum=sum+add;
    }
    printf("%d+%d%d+...+%d=%d.\n",a,a,a,(int)(temp),sum);
}

收到的鲜花

• shinobifc 于 2010-02-08 21:21 送鲜花  1朵   附言：我很赞同

#include <stdio.h>

int main (void)
{
    char a,n;
    long  temp=0, sum=0;
    int i;

    printf("Please input positive integers 'a' and 'n':\n");
    scanf("%c %c", &a, &n);
    while (a<'1'  || n<'1' || a > '9' || n > '9')
    {
        printf("Sorry! 'a' or 'n' doesn't fit. Please input 'a' & 'n' again:\n");
        scanf("%c %c", &a, &n);
    }

    a-=48;
    n-=48;

    for (i=0; i<n; i++)
    {
        temp = temp*10 + a;
        sum += temp;
    }

    if (n == 1)
    {
        printf("%d=%d.\n",a,a);
    }
    else if (n == 2)
    {
        printf("%d+%ld=%ld.\n",a, temp ,sum);
    }
    else if (n == 3)
    {
        printf("%d+%d%d+%ld=%ld.\n",a, a,a, temp, sum);
    }
    else if (n > 3)
    {

        printf("%d+%d%d+...+%ld=%ld.\n",a,a,a,temp,sum);
    }

    return 0;
}

收到的鲜花

• shinobifc 于 2010-02-08 21:21 送鲜花  1朵   附言：原创内容

那该如何解决呢？

实际上是不需要幂函数的，你考虑一下前后项的关系。
#include <stdio.h>
main ()
{
    long int temp=0, sum=0;
    int a, n, i;

    printf("Please input positive integers 'a' and 'n':\n");
    scanf("%d%d", &a, &n);

    while (a<=0||a>=10 || n<=0)
    {
        printf("Sorry! 'a' or 'n' doesn't fit. Please input 'a' & 'n' again:\n");
        scanf("%d%d", &a, &n);
    }

    for (i=0; i<n; i++)
    {
        temp=temp*10+a;
        sum+=temp;
    }

    printf("%d+%d%d+...+%ld=%ld.\n",a,a,a,temp,sum);
}

方法有很多，最简单的就是你自己写一个pow函数，返回值为long
如long myPow(int k) {
int i = 0;
long temp = 1;

for(i = 0; i < k; i++);
    temp *= 10;
}

我稍微改了下能运行（我认为要注意while 与 if 的区别）
#include <math.h>
#include<stdio.h>
main ()
{
    long int a, temp=0, add=0, sum=0;
    int n, i;
    printf("Please input positive integers 'a' and 'n':\n");
    scanf("%ld%d",&a,&n);
    if(a<=0 || n<=0 || a>=10)
    {
        printf("Sorry! 'a' or 'n' doesn't fit. Please input 'a' & 'n' again:\n");
        scanf("%ld%d",&a,&n);
    }
    for (i=n;i>0; i--)
    {
        temp=temp+a*pow(10,i-1);
    }
    add=(int)(temp);
    sum=add;
    while (add>9)
    {
        add=(int)(add/10);
        sum=sum+add;
    }
    printf("%d+%d%d+...+%d=%d.\n",a,a,a,(int)(temp),sum);
}

#include <math.h>
#include<stdio.h>
main () {
    long int a, j=1, add=0;
    int i,n;
    printf("Please input positive integers 'a' and 'n':\n");
    scanf("%ld%d",&a,&n);
    for (i=n; i>0; i--) {
       add+=a*Mypow(i)*j;
       j++;
    }
    printf("%ld\n",add);
    getch();

}
int Mypow(int n) {
    int i,k=1;
    for (i=1; i<n; i++)
        k*=10;
    return (k);
}

tc下写的，  没有加判断a n 的语句  只有简单的算法！！