First, let give the bits some names, I put them inside the parentheses:
Petal 1 (p1) = II (c2) + III (c3) + IV (c4)
Petal 2 (p2) = I (c1) + III (c3) + IV (c4)
Petal 3 (p3) = I (c1) + II (c2) + IV (c4)

As p1, p2 and p3 are created from c1, c2, c3, c4, we check each change in c1, c2, c3, c4:
When c1 change: p2 and p3 won't match the value calculated using the formular above (i.e: c1 + c3 + c4 != p2 and c1 + c2 + c4 != p3)
When c2 change: p1 and p3 will go wrong
When c3 change: p1 and p2 will go wrong
As c4 contribute to the value of all three petals, if it is changed then p1, p2 and p3 will all go wrong.

In short:
Calculate ep1, ep2 and ep3 from the first 4 bits (the e in ep stand for expected), then compare them with the given p1, p2 and p3:
    All three pairs do not match: c4 is changed
    p1 and p2 do not match:       c3 is changed
    p1 and p3 do not match:       c2 is changed
    p2 and p3 do not match:       c1 is changed
    Only p1 not match:            p1 is changed
    Only p2 not match:            p2 is changed
    Only p3 not match:            p3 is changed

Instead of "if you enemy..." it should be "if your enemy...".

I have used bitmasks.
There are only 16 possible Hamming codes.
So,  I have precalculated.
Then check,  whether input is a valid code.
Then try to find such valid code that it differs from input in exactly one position (I have
used xoring for that)  and print that codr.  Don't forget about zero code.  That is a special case.

if((a[2]+a[3]+a[4])%2==a[5] && (a[1]+a[3]+a[4])%2==a[6] && (a[2]+a[1]+a[4])%2==a[7])
shu bajarilguncha ketma-ketlikni o'zgartirib borasiz 0 bo'lsa 1 va aksincha 1 bo'lsa 0
Masalan:
0 1 0 1 1 0 1  for petal 0 1 0
1 1 0 1 1 0 1  for petal 0 0 1
0 0 0 1 1 0 1  for petal 1 1 1
0 1 1 1 1 0 1  for petal 1 0 0
0 1 0 0 1 0 1  for petal 1 0 1

#include<stdio.h>
#include<stdlib.h>



int main()
{
int a[8],i,q1=0,q2=1,q3=2,q4=3,r1=4,r2=5,r3=6;
for(i=0;i<7;i++)
{
scanf("%d",&a[i]);
}

if((a[q2]+a[q3]+a[q4])%2==a[r1]&&(a[q1]+a[q3]+a[q4])%2==a[r2]&&(a[q1]+a[q2]+a[q4])%2==a[r3])
{
for(i=0;i<7;i++)
{
printf("%d",a[i]);
printf(" ");
}
}
else if ((a[q2]+a[q3]+a[q4])%2!=a[r1]&&(a[q1]+a[q3]+a[q4])%2!=a[r2]&&(a[q1]+a[q2]+a[q4])%2==a[r3])
{
if(a[q3]==1)
{
a[q3]=0;
}
else
{
a[q3]=1;
}
for(i=0;i<7;i++)
{
printf("%d",a[i]);
printf(" ");
}
}
else if((a[q2]+a[q3]+a[q4])%2!=a[r1]&&(a[q1]+a[q3]+a[q4])%2==a[r2]&&(a[q1]+a[q2]+a[q4])%2!=a[r3])
{
if(a[q2]==1)
{
a[q2]=0;
}
else
{
a[q2]=1;
}
for(i=0;i<7;i++)
{
printf("%d",a[i]);
printf(" ");
}
}
else if((a[q2]+a[q3]+a[q4])%2==a[r1]&&(a[q1]+a[q3]+a[q4])%2!=a[r2]&&(a[q1]+a[q2]+a[q4])%2!=a[r3])
{
if(a[q1]==0)
{
a[q1]=1;
}
else
{
a[q1]=0;
}
for(i=0;i<7;i++)
{
printf("%d",a[i]);
printf(" ");
}
}
else
{
if(a[q4]==1)
{
a[q4]=0;
}
else
{
a[q4]=1;
}
for(i=0;i<7;i++)
{
printf("%d",a[i]);
printf(" ");
}
}

return 0;
}

IN
0 1 1 0 0 0 0
OUT
1 1 1 0 0 0 0

IN
1 0 1 0 0 0 0
OUT
1 1 1 0 0 0 0

IN
1 1 0 0 0 0 0
OUT
1 1 1 0 0 0 0

IN
1 1 1 1 0 0 0
OUT
1 1 1 0 0 0 0

IN
1 1 1 0 1 0 0
OUT
1 1 1 0 0 0 0

IN
1 1 1 0 0 1 0
OUT
1 1 1 0 0 0 0

IN
1 1 1 0 0 0 1
OUT
1 1 1 0 0 0 0