A positive integer x, present m more than a; Present more than n b, m, n, a, b, x
Note: x is probably not the solution, if there is no solution, is given no prompt solutions

CodePudding user response:

Not told you thinking?
X=a=I + n * m * j + b
Launch
J=n * m * I + (a - b)
Launch
J=(m * I + (a - b))/n
Because I and j is an integer (m * I + (a - b)) must can ensure j can be divided exactly by n is integer
So, I have been cycle to a maximum of integer from 0, see if you can have a meet the conditions of I, find I is a solution, otherwise it is no solution

 int mian () {
Int a, b, m, n, I; 
The scanf (" % d % d % d % d ", & amp; M, & amp; N, & amp; A, & amp; B); 
for (i=0; ; I++) {
If (i<0 {
Printf (" no solution "); 
break;//I turned negative seams show unanswered 
} 
Else if ((m * I + a - b) % n==0) {
Printf (" has a solution: x=% d ", m * I + a); 
break; 
} 
} 
return 0; 
} 

CodePudding user response:

Simple and crude, only supplies the reference:

 # include & lt; stdio.h> 

Int main (void) 
{
Unsigned int a, b, m, n, x. 
The scanf (" % d % d % d % d ", & amp; M, & amp; N, & amp; A, & amp; B); 

X=1; 
While (x) 
{
If (m=(x %=a) & amp; & (x % n==b)) 
{
Printf (" has a solution: x=% u ", x); 
return 0; 
} 
x++; 
} 

Printf (" no solution "); 

return 1; 
} 

CodePudding user response:

Because you have the wrong ah, ha ha ha ha

CodePudding user response:

The

reference 4 floor MSWPP reply:

because you have the wrong ah, ha ha ha ha

Stare big eyes, all is right,