Thursday, August 03, 2006

Intelligence Formula using for loop.

According to a study, the approximate level of intelligence of a person can be calculated
using the formula:

i=2 + (y+0.5x)

Write a program which will produce a table of values of i, y and x, where y varies from 1 to 6,
and, for each value of y, x varies from 5.5 to 12.5 in steps of 0.5 .





#include<stdio.h>
main()
{


int y;
float i,x;

for (y=1; y<=6; y++)

{
for (x=5.5; x<=12.5;x+=0.5)

{
i=2 + (y+ (0.5*x)) ;


printf("\n i=%f, y=%d, x=%f", i, y, x);
continue;
}
}

}



The file can be found at:
Download File

3 comments:

Anonymous said...

Why the use of continue?

yellaling said...

try this code its not wrking
#include
#include
int main()
{
float i,x,y;
//x=5.5;
int j;
for(j=1;j<=6;j++)
{
for(x=5.5;x<=12.5;x+=5.5)
{
i=2+(j+0.5x);
//x=x+0.5;
printf("%f\n",i);
continue;
}

}
getch();
return 0;
}

Anonymous said...

main()
{
float i,x;
int j;
for(j=1;j<=6;j++)
{
for(x=5.5;x<=12.5;x+=0.5) //x=x+0.5;
{
i=2+(j+0.5*x);

printf("%f %d %f \n",i, j, x);
continue;
}

}
getch();
return 0;
}
Now code will work