【c语言编程题】输入a,n输出s=a+aa+aaa+…n项之和。要求定义一个fn(a,n)函数实现
#include
#include
int main()
{
int a,n;
int i;
long sum = 0;
int m = 0;
printf("Input a,n:
");
scanf("%d,%d",&a,&n);
for(i=1; i<=n; i++)
{
m = a*pow(10,i-1) + m;
sum = sum + m;
}
printf("sum=%ld
",sum);
return 0;
}
答案可以过作业
在CB17.12中有可能出现 当i=3时 pow（10，i-1）为99的情况，求大神解答一下
这个题利用循环结构就好了，具体代码如下：
#include #include#include int sum(int n, int a); main(){ int a , n ;printf("Input a,n:
");scanf("%d,%d",&n,&a);printf("sum=%ld
",sum(n,a));system("pause"); } int sum(int a,int n) {int i ;int A=a ;int s=0 ;for(i=1;i<=n;i++){s+=A;A=A+a*pow((double)10,i);}return s; }
#include<stdio.h>
int main()
{
int a,n,s=0;
int fn(int a,int n);
while(1)
{
printf("Please enter a positive integer(0~9):a=");
scanf("%d",&a);
if(a>=0 && a<=9) break;
}
printf("Please enter a positive integer:n=");
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
s+=fn(a,i);
}
printf("s=%d",s);
}
int fn(int a,int n)
{
int t;
t=a;
for(int i=2;i<=n;i++)
{
a=a*10+t;
}
return a;
}
#include<stdio.h>
#include<math.h>
int fn(an,nn)
{
int in;
long int sn=0;
for (in=1; in<=i;i++)
{
sn=sn+an*10^(in-1);
}
}
void main(void)
{
int n;
int a;
int i;
int s=0;
printf("input a n");
scanf ("%d %d",&a,&n);
for(i=1;i<=n;i++)
{
s=s+fn(a,i)
}
}
#include <stdio.h>
long fn(int a, int n);
int main(int argc, char argv)
{
int a = 0, n = 0;
printf(" Please input 2 number(a,n): ");
scanf("%d,%d", &a, &n);
printf("
a=%d, n=%d sun = %d
", a, n, fn(a, n));
return 0;
}
long fn(int a, int n)
{
int i;
long sun = 0, rat = 1;
for ( i = 0; n - i > 0; i++)
{
sun += a * rat * (n-i);
rat *= 10;
}
return sun;
}
S等于二分之N方加上二分之一N 然后整体乘以a
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