# Learning Log No. 6

1. By programming, count the number of 9 in 1~n
```#include <stdio.h>

void fun(int n)
{
int i;
int temp;
int count = 0;
int b;

for(i = 1;i <= n;i++)
{
temp = i;
while(temp != 0)
{
b=temp % 10;
temp=temp / 10;
if(b == 9)
{
count++;
}
}
}

printf("The result is: %d\n",count);
}

int main()
{
int n;

printf("Enter a number:\n");
scanf("%d",&n);

fun(n);

return 0;
}
```

2. There are n people in a circle, numbering in sequence, starting with the first report (from 1 to 3). Everyone who reports to 3 quits the circle and asks the last one who left behind the original number.

```#include <stdio.h>

int main()
{
int n;
printf("Enter number of people:\n");
scanf("%d",&n);
int a[1000];
int i,temp;
int count = 0;
temp = n;
for(i = 0;i < n;i++)
{
a[i] = i+1;
}
i = 0;
while (n>1)
{
if(a[i] != 0)
{
count++;
}
if(count == 3)
{
a[i] = 0;
count = 0;
n--;
}
i++;
if(i == temp)
{
i = 0;
}
}
for(i = 0;i < temp;i++)
{
if(a[i] != 0)
{
printf("The rest are:%d Number\n",a[i]);
}
}
return 0;
}
```

3. Enter 5 numbers (negative, decimal) to arrange them in order from smallest to largest

```#include <stdio.h>

float  paixu(float b[5]);
int main()
{
float a[5];
int i;

for(i = 0;i < 5;i++)
scanf("%f",&a[i]);
paixu(a);

printf("After sorting, the result is:\n");
for(i = 0;i < 5;i++)
printf("%2.1f\t",a[i]);
printf("\n");

return 0;
}

float  paixu(float b[])
{
int i, j;
float t = 0.0;

for(i = 0;i < 5;i++)
{
for(j = i + 1; j < 5;j++)
{
if(b[j] < b[i])
{
t = b[i];
b[i] = b[j];
b[j] = t;
}
}
}
}
```

4. Find prime numbers within 100 and print them all out

```#include <stdio.h>

int main()
{
int i,j;
int temp;

for(i = 2;i <=100;i++)
{
int temp = 1;
for(j = 2; j < i;j++)
{
if(i % j == 0)
{
temp = 0;
break;
}
}
if(temp == 1)
{
printf("%d ",i);
}
}
return 0;
}
```

5. If a number is exactly equal to the sum of its factors, it is called a "complete number". For example: 6=1+2+3. Program to find the complete number within 1000

```#include<stdio.h>

int main()
{
printf("1000 Number of completions within: \n");
int i = 0, j = 0, sum = 0;
for(i = 1; i <= 1000; ++i)
{
for(j = 1; j < i; ++j)
{
if(0 == i % j)
{
sum = sum + j;
}
}

if(sum == i)
{
printf("%d ",i);

}
sum =0;
}
printf("\n");
return 0;
}
```

Tags: Programming REST

Posted on Fri, 14 Feb 2020 08:34:30 -0800 by jtapoling