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C언어/과제

C언어 과제 07

1. 문제에 나온 그림대로 출력하세요. (중첩 for문 사용)

1)

      *

         *

            *

               *

                  *

#include <stdio.h>

 

int main()

{

 

           int cnum=0;

           int count;

           int count1;

          

           for(count=0; 5 > count ;count++)

           {

                     for(count1=0; cnum>=count1; count1++)

                     {

                                printf("%c",32);

                     }

                     printf("*\n");

                     cnum++;

           }

           return 0;

}

2)

                 *

              *

           *

        *

     *

#include <stdio.h>

 

int main()

{

 

           int cnum=5;

           int count;

           int count1;

          

           for(count=0; 5 > count ;count++)

           {

                     for(count1=0; cnum>=count1; count1++)

                     {

                                printf("%c",32);

                     }

                     printf("*\n");

                     --cnum;

           }

           return 0;

}

2. 출력할 줄의 개수를 입력받은 후 직삼각형을 다음과 같이 출력하세요. (중첩 for문 사용)

줄 개수 입력 : 5

*

* *

* * *

* * * *

* * * * *

#include <stdio.h>

 

int main()

{

 

           int cnum=0;

           int count;

           int count1;

          

           for(count=0; 5 > count ;count++)

           {

                     for(count1=0; cnum>=count1; count1++)

                     {

                                printf("*");

                     }

                     printf("\n");

                     cnum++;

           }

           return 0;

}

3. 다음과 같은 형태로 문자열이 출력되는 프로그램을 작성하시오. (중첩 for문 사용)

1)

           A

           A         B

           A         B         C

           A         B         C         D

           A         B         C         D         E

#include <stdio.h>

 

int main()

{

 

           char ch[5]="ABCDE";

           int count;

           int chcnt;

          

           for(count=0; 4>=count; count++)

           {

                     for(chcnt=0; count>=chcnt; chcnt++)

                     {

                                printf("%c", ch[chcnt]);

                     }

                     printf("\n");

           }

           return 0;

}

2)                                                  a

                                          b         a

                                c         b         a

                     d         c         b         a

           e         d         c         b         a

#include <stdio.h>

 

int main()

{

 

             char ch[5]="ABCDE";

             int count;

             int chcnt=0;

             int cnt;

 

             for(count=4; 0<=count; count--)

             {

                           for(cnt=1; cnt<=count ;cnt++)

                           {

                                        printf(" ");

                           }

                           for(cnt=chcnt; 0<=cnt; cnt--)

                           {

                                        printf("%c", ch[cnt]);

                           }

                           chcnt++;

                           printf("\n");

             }

             return 0;

}

3. 문제에 나온 그림대로 출력하세요. (중첩 for문 사용)

      *           *

        *       *

          *   *

            *

          *   *

        *       *

      *           *

#include <stdio.h>

 

int main()

{

        int cnt;

        int cnt1=5;

        int i,k;

 

        for(cnt=1; 4>=cnt; cnt++)

        {

                for(i=1; i<cnt; i++)

                {

                        printf(" ");

                }

                if(4>cnt)

                {

                        printf("*");

                }

                for(k=1; k<=cnt1; k++)

                {

                        printf(" ");

                }

                printf("*\n");

                cnt1=cnt1-2;

        }

        cnt1=1;

        for(cnt=1; 3>=cnt; cnt++)

        {

                for(i=2; i>=cnt; i--)

                {

                        printf(" ");

                }

                printf("*");

                for(k=1; k<=cnt1; k++)

                {

                        printf(" ");

                }

                printf("*\n");

                cnt1=cnt1+2;

        }

        return 0;

}

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