Output less the first number

/* if an n is a positive integer is equal to its n number of n to the power and the number of known as n the power since the number, the power design for 3 ~ 6 from number, 
* * output format requirements: "% d bits from the exponential are:" "% ld \ t" "\ n" (after each newline) 
Application example are as follows: 
3 since the power number: 153 370 371 407 
4 since the exponential are: 1634 8208 9474 
5 the power number: 54748 92727 93084 
Six since the power number: 548834 */
# include 
# include 
Int HHH (int n) 
{
Long a [10], I, j, x, y, t, sum=0; 
X=pow (10, n - 1); 
Y=pow (10, n); 
For (I=x; i{
T=I; 
for(j=0; J{
A [j]=t % 10; 
T/t=10; 
} 
for(j=0; J{
Sum +=pow (a [j], n); 
} 
If (sum==I) 
{
Printf (" % ld \ t ", sum); 
} 
Sum=0; 
} 
} 
Int main (void) 
{
int i; 
For (I=3; i<=6; I++) 
{
Printf (" % d a since the exponential are: ", I); 
HHH (I); 
printf("\n"); 
} 
} 

The running result is like this:

CodePudding user response:

Didn't see you run, I run, there is no problem:
The int HHH () function is slightly changed:

 int HHH (int n) 
{
Long a [10], I, j, x, y, t, sum; 
X=pow (10, n - 1); 
Y=pow (10, n); 
For (I=x; i{
T=I; 
Sum=0;//the sum initialization here 
for(j=0; J{
A [j]=t % 10; 
T/t=10; 
} 
for(j=0; J{
Sum +=pow (a [j], n); 
} 
If (sum==I) 
{
Printf (" % ld \ t ", sum);//"% ld \ t" join a space in between. 
} 
} 
}