8/29/2011
f(N) counts number of '1'. For example f(13) = 6. So f(1)=1. What next number do satisfy the rule?
f(N) counts number of '1'. For example f(13) = 6. So f(1)=1. What next number do satisfy the rule?
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양수 n에 대해서 1과 n 사이에 1이 나오는 횟수를 나타내는 함수를 f(n)이라고 한다. 예를 들어 f(13)=6이다. f(n)=n이 되는 첫번째 양수는 1이다. 두번째 양수는 무엇인가?
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This is my code.
ReplyDeleteLet's compare to each code.
Bye Bye~ ^^
int CountOne(int input)
{
int count =0;
while(1)
{
if( input%10 == 1)
count++;
input = input/10;
if(input == 0)
break;
}
return count;
}
int main()
{
int i=2;
int count=1;
while(1)
{
count += CountOne(i);
if( i == count)
break;
i=i+1;
}
printf("f(%d) = %d\n", i, count);
return 0;
}
JeKang's code:
ReplyDelete#include
void main()
{
unsigned long Num, End, Temp;
Num=0;
End=0;
while(Num<=1 || Num!=End){
Num+=1;
Temp=Num;
while(Temp!=0){
if(Temp%10==1) End+=1;
Temp=(unsigned long)(Temp/10);
}
}
printf("입력값 Num = %d\n결과값 End = %d\n",Num,End);
}
제강아 소스 잘 짰네..
ReplyDelete내가 볼때는, while문 조건이 조금 복잡해서 나중에 보면 이해가 쉽게 안갈 것 같고..
while문 안에 while문은 특정 일만 하니깐 함수로 만들면 소스가 더 간결해 질 것 같다.
잘했네..
Good job~!!
JeKang :
ReplyDelete네 함수로도 만들어 볼께요. 감사합니다~
민현규
ReplyDelete#include
#include
int fun(char* a);
void main()
{
int d;
int p=0;
int i=0;
char a[20]={0,};
printf("숫자입력하세요:");
scanf("%d",&d);
for(i=0;i<=d;i++)
{
itoa(i,a,10);
p=p+fun(a);
if(p==i)
printf("%d\n",p);
}
}
int fun(char* a)
{
int i=0;
int count=0;
for(i=0;i<= 20 ;i++)
{
if(a[i]=='1') count+=1;
}
return count;
}