let say i have two loops and i want to check if the varibles from one loop can evenly divided like this
how can i make the(x%y==0) condition to be checked for every y condtion ranging fromCode:for (y=1;y<5;y++){ for (x=0;x<1000;x++){ if (x%y==0){ printf("this one: %d \n",x)}}}
1 to 5
i mean how can it make it find every number that can be eveny divided to all number ranging from 1 to 5
thanks
Presumably by putting a semicolon ";" after your printf statement. Once you have that syntactically correct, your if statement will check 4000 different cases (y 1,2,3,4; x from 0 to 999).
If i understand you question correctly this should do itCode:for (y=1;y<6;y++){ for (x=0;x<1000;x++){ if (x%y==0){ printf("this one: %d \n",x); } } }
edit:
as stated above, this will not check x=1000
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Maybe you want something along these lines:
Code:#include <stdio.h> int main(void) { int x,y; for (x=0;x<1000;x++) { int flag = 0; for (y=1;y<=5;y++) { if (x%y) { flag = 1; /*found y that is not dividable by */ break; } } if(flag == 0) { /* x is dividable by all y */ printf("this one: %d \n",x); } } return 0; }
Last edited by vart; 03-28-2009 at 02:06 PM.
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler
