How to print 1 ＋ 11 ＋ 111 ＋ 1111 + …… ＋ 111111111 =?

血染菊花

I have a question to ask:
Ask to print:
1 + 11 + 111 + 1111 + ... + 111111111 =?
2 + 22 + 222 + 2222 + ... + 222222222 =?
3 + 33 + 333 + 3333 + ... + 333333333 =?
...
...
9 + 99 + 999 + 9999 + ... + 999999999 =?

The code I wrote is as follows:
main ()
{
int i, j;
long f1, f2, sum = 0;
for (i = 1; i <= 9; i ++) {
f1 = i;
printf ("sum =");
for (j = 1; j <= 9; j ++) {
f2 = f1 * 10 + i;
sum + = f1;
if (j == 9) printf ("% d =", f1);
else printf ("% d +", f1);
f1 = f2;
}
printf ("% d\n", sum);
}
}

I wonder what went wrong? Please help me!

暴走的熊猫

The data is too big to overflow

sjtwenzhou

Note the statement inside the first loop: sum = 0;
Otherwise it will add up

sjtwenzhou

#include <stdio.h>
int main ()
{
int i, j;
long f1, f2, sum = 0;
for (i = 1; i <= 9; i ++)
{
f1 = i;
sum = 0; // plus
printf ("sum =");
for (j = 1; j <= 9; j ++)
{
f2 = f1 * 10 + i;
sum + = f1;
if (j == 9) printf ("% d =", f1);
else printf ("% d +", f1);
f1 = f2;
}
printf ("% d\n", sum);
}
return 0;
}

血染菊花

Still not working, it seems like data overflow, how to do it?

血染菊花

What does the last return 0; mean? Thank you!

帝王至尊

Returns a number to the main function,

帝王至尊

#include <stdio.h>
int main ()
{
int i, j;
long int f1, f2, sum = 0;
for (i = 1; i <= 9; i ++)
{
f1 = i;
sum = 0;
printf ("sum =");
for (j = 1; j <= 9; j ++)
{
f2 = f1 * 10 + i;
sum + = f1;
if (j == 9) printf ("% ld =", f1);
else printf ("% ld +", f1);
f1 = f2;
}
printf ("% ld\n", sum);
}
return 0;
}
Improved, that is, all the numbers are output using ld long shaping