11-01 Exercise 3: Fibonacci numbers

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In mathematics, the Fibonacci numbers or Fibonacci sequence are the numbers in the following integer sequence: 0,1,1,2,3,5,8,13,21,34,55,89,144

前13的斐波那契數列：0,1,1,2,3,5,8,13,21,34,55,89,144

By definition, the first two numbers in the Fibonacci sequence is 0 and 1, and each subsequent number is the sum of the previous two. In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the relation Fn = Fn-1 + Fn-2 with seed values F0 = 0, F1= 1

斐波那契數由0和1開始，其後的數就由之前的兩個數相加。用數學表示 Fn = (Fn-1) + (Fn-2)，F0 = 0, F1 = 1

Write a program that asks the user to type an integer n, then display Fn on the screen.

寫一個輸入一個整數n，輸出第n個斐波那契數的程式

Example input 1:

5

Example output 1:

5

 

Example input 2:

12

Example output 2:

144

 

[Exercise] Coding (C) - asked in Chapter 6: Loops by (5.2k points)
ID: 29830 - Available when: 2017-11-02 18:00 - Due to: Unlimited
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Case 0: Correct output
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prog.c: In function 'main':
prog.c:6:10: warning: null argument where non-null required (argument 1) [-Wnonnull]
 if(a==0) printf(0);
          ^~~~~~
prog.c:7:22: warning: passing argument 1 of 'printf' makes pointer from integer without a cast [-Wint-conversion]
 else if(a==1) printf(1);
                      ^
In file included from prog.c:1:0:
/usr/include/stdio.h:364:12: note: expected 'const char * restrict' but argument is of type 'int'
 extern int printf (const char *__restrict __format, ...);
            ^~~~~~
prog.c:7:1: warning: format not a string literal and no format arguments [-Wformat-security]
 else if(a==1) printf(1);
 ^~~~
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answered by (-254 points)
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prog.c: In function 'main':
prog.c:9:10: warning: null argument where non-null required (argument 1) [-Wnonnull]
 if(a==0) printf(0);
          ^~~~~~
prog.c:10:22: warning: passing argument 1 of 'printf' makes pointer from integer without a cast [-Wint-conversion]
 else if(a==1) printf(1);
                      ^
In file included from prog.c:1:0:
/usr/include/stdio.h:364:12: note: expected 'const char * restrict' but argument is of type 'int'
 extern int printf (const char *__restrict __format, ...);
            ^~~~~~
prog.c:10:1: warning: format not a string literal and no format arguments [-Wformat-security]
 else if(a==1) printf(1);
 ^~~~
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answered by (-107 points)
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Case 1: Correct output
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answered by (-107 points)
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prog.c: In function 'main':
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          ^~~~~~
prog.c:10:22: warning: passing argument 1 of 'printf' makes pointer from integer without a cast [-Wint-conversion]
 else if(i==1) printf(1);
                      ^
In file included from prog.c:1:0:
/usr/include/stdio.h:364:12: note: expected 'const char * restrict' but argument is of type 'int'
 extern int printf (const char *__restrict __format, ...);
            ^~~~~~
prog.c:10:1: warning: format not a string literal and no format arguments [-Wformat-security]
 else if(i==1) printf(1);
 ^~~~
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answered by (-254 points)
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answered by (-254 points)
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