Required to achieve a simple calculation of integral factor and function, and another function, using the output of two positive integers m and n (0 & lt; M n 10000) or less or less between all the number, the number is the sum of the number in addition to its own equals the factor, for example: 6=1 + 2 + 3, among them 1, 2, 3 to 6 factors,

How to find a common factor simpler, cyclic code too many times to handle long

 # include & lt; stdio.h> 

Int factorsum (int number); 
M void PrintPN (int, int n); 

Int main () 
{
Int m, n; 

The scanf (" % d % d ", & amp; M, & amp; N); 
If (factorsum (m)==m) printf (" % d is a perfect number \ n ", m); 
If (factorsum (n)==n) printf (" % d is a perfect number \ n ", n); 
PrintPN (m, n); 

return 0; 
} 

Int factorsum (int number)//determine whether to the number of 
{
int i,j; 
Int m=1; 
Int n, num=0; 
Int a [100]={0}; 
A [0]=1; 
If (number==0) 
return -1; 
If (number==1) 
return 1; 
For (I=2; I & lt; Number; I + +) 
{
For (j=I + 1; J & lt; Number; J + +) 
{
If (I * j==number) 
{
A [m]=I; 
M + +; 
A [m]=j; 
M + +; 
} 

} 
} 
For (n=0; N & lt; 100; N + +) 
{
If (a [n]==0) 
break; 
Num=num + a, [n]. 
} 
If (num==number) 
return num; 
The else 
return 0; 

} 

Void PrintPN (int m, int n)//find out between m and n number 
{
Int q; 
int i,j; 
int num=0; 

For (int k=m; k <=n; K++) 
{
If (factorsum (k)==k) 
{
Num=1; 
Int p=1; 
Int a [100]={0}; 
A [0]=1; 
For (I=2; I & lt; k; I + +) 
{
For (j=I + 1; J & lt; k; J + +) 
{
If (I * j)==k 
{
A [p]=I; 
P + +; 
A [p]=j; 
P + +; 

} 
} 
} 
//order 
Int b; 
For (b=0; [b]!=0; B++) 
{
For (int c=b + 1; A [c]!=0; C + +) 
{
If (a [b] & gt; A [c]) 
{
Int the key; 
Key=a [b]; 
A [b]=a [c]; 
A [c]=key; 
} 

} 

} 
Printf (" % d=", k); 
For (q=0; Q & lt; 100; Q + +) 
{
If (a [q]==0) 
{
printf("\n"); 
break; 
} 
If (q==0) 
Printf (" % d ", a [q]); 
The else 
Printf (" + % d ", a [q]); 

} 
} 
} 
If (num==0) 
Printf (" No perfect number \ n "); 

} 

CodePudding user response:

Random search code of positive integer decomposition factor

CodePudding user response:

The original poster is too complex, for reference:

 # include & lt; stdio.h> 
Int factorsum (int n);//determine whether to the number of 

Int main () 
{
Int m, n; 
The scanf (" % d % d ", & amp; M, & amp; N); 
If (m<=0 | | n> 10000) return 1; 

For (int I=m; IIf (factorsum (I)) printf (" % d is a perfect number \ n ", I); 
} 

return 0; 
} 


Int factorsum (int n) 
{
Int k=1, s=0; 
While (k & lt;=n/2) {
If (n % k==0) s +=k; 
k++; 
} 
If (s==n) return 1; 
return 0; 
} 

CodePudding user response:

Fixed line 10 upstairs:

 for (int I=m; I<=n; I++) {

CodePudding user response:

 # include & lt; stdio.h> 

Int factorsum (int number); 
M void PrintPN (int, int n); 

The static void swap (int * a, int * b); 
The static void print_perfect (int m, int n); 

Int main () 
{
Int m, n; 

The scanf (" % d % d ", & amp; M, & amp; N); 

If (m & gt; N) {
Swap (& amp; M, & amp; N); 
} 

Print_perfect (m, n); 
/* 
If (factorsum (m)==m) printf (" % d is a perfect number \ n ", m); 
If (factorsum (n)==n) printf (" % d is a perfect number \ n ", n); 
PrintPN (m, n); 
*/
return 0; 
} 

The static void swap (int * a, int * b) 
{
Int TMP. 

TMP=* a; 
*=* b; 
* b=TMP; 
} 

The static void print_perfect (int m, int n) 
{
int i; 

For (I=m; I & lt; n; I++) {
If (I==1) 
continue; 
{if (factorsum (I)) 
Printf (" % d is a perfect number! \ n ", I); 
} 
} 
} 


Int factorsum (int number)//determine whether to the number of 
{
int i; 
Int sum=1; 


For (I=2; I & lt; Number/2 + 1; I++) {
If (number % I==0) 
sum +=i; 
} 
If (sum==number) 
return 1; 

return 0; 
/* 
int i,j; 
Int m=1; 
Int n, num=0; 
Int a [100]={0}; 
A [0]=1; 
If (number==0) 
return -1; 
If (number==1) 
return 1; 
For (I=2; I & lt; Number; I + +) 
{
For (j=I + 1; J & lt; Number; J + +) 
{
If (I * j==number) 
{
A [m]=I; 
M + +; 
A [m]=j; 
M + +; 
} 

} 
} 
For (n=0; N & lt; 100; N + +) 
{
If (a [n]==0) 
break; 
Num=num + a, [n]. 
} 
If (num==number) 
return num; 
The else 
return 0; 
*/
} 

Void PrintPN (int m, int n)//find out between m and n number 
{
Int q; 
int i,j; 
int num=0; 

For (int k=m; k <=n; K++) 
{
If (factorsum (k)==k) 
{
Num=1; 
Int p=1; 
Int a [100]={0}; 
A [0]=1; 
For (I=2; I & lt; k; I + +) 
{
For (j=I + 1; J & lt; k; J + +) 
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